Permutations and Combinations
When given a number of objects and we would to find the possible different arrangements or selections , we may use permutations or combinations.
Permutations deal with the arrangement of objects. The order of the objects is important. The number of different arrangements of r objects from n distinct object is,
\(\\[20pt]\hspace{2em} ^{n}P_{r } \ = \ \displaystyle \frac{n!}{(n-r)!} \)
Combinations deal with the selection of objects. The order of the objects is not important. The number of different selections of r objects from n distinct object is,
\(\\[20pt]\hspace{2em} ^{n}C_{r } \ = \ \displaystyle \binom{n}{r} \ = \ \frac{n!}{r!(n-r)!} \)
Notice that it is also used as the binomial coefficients in Binomial Expansion and Binomial Series.
Let’s see a simple example below to illustrate the difference between permutations and combinations.
We have 3 letters, A, B and C and we would like to find the permutations and combinations of 2 letters from the 3 letters.
For permutations, we have the following arrangements: AB, BA, AC, CA, BC and CB.
For combinations, we have the following selections: AB, AC and BC.
As you can see, AB and BA, AC and CA or BC and CB are considered as the same selection since the order of the objects is not important.
In other words, combinations is actually the permutations of r identical objects from n distinct object.
Try the exercises below and if you need any help, just look at the solution I have written. Cheers ! =) .
\(\\[1pt]\)
EXERCISE 5A:
\({\small 1.\enspace}\) Seven different cars are to be loaded on to a transporter truck. In how many different ways can the cars be arranged?
The arrangements of 7 different cars are the permutations of 7 objects from 7 distinct objects, thus:
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{7}P_{7} \ = \ {\large \frac{7!}{(7 \ – \ 7)!} } }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{7!}{0!} } }\)
\(\\[10pt]\hspace{4em}{\small = \ 7 \ \times \ 6 \ \times \ 5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1}\)
\(\\[10pt]\hspace{4em}{\small = \ 5040 }\)
There are 5040 ways to arrange the cars.
\(\\[1pt]\)
\({\small 2.\enspace}\) How many numbers are there between 1245 and 5421 inclusive which contain each of the digits 1, 2, 4 and 5 once and once only?
There are 4 distinct digits: 1, 2, 4, 5.
Since both 1245 and 5421 are already the smallest and largest possible number respectively, the number of permutations is:
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{4}P_{4} \ = \ {\large \frac{4!}{(4 \ – \ 4)!} } }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{4!}{0!} } }\)
\(\\[10pt]\hspace{4em}{\small = \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1}\)
\(\\[10pt]\hspace{4em}{\small = \ 24 }\)
There are 24 numbers can be arranged between 1245 and 5421.
\(\\[1pt]\)
\({\small 3.\enspace}\) An artist is going to arrange five paintings in a row on a wall. In how many ways can this be done?
The arrangements of 5 different paintings are the permutations of 5 objects from 5 distinct objects, thus:
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{5}P_{5} \ = \ {\large \frac{5!}{(5 \ – \ 5)!} } }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{5!}{0!} } }\)
\(\\[10pt]\hspace{4em}{\small = \ 5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1}\)
\(\\[10pt]\hspace{4em}{\small = \ 120 }\)
There are 120 ways to arrange the paintings.
\(\\[1pt]\)
\({\small 4.\enspace}\) Ten athletes are running in a 100-metre race. In how many different ways can the first three places be filled?
There are 10 athletes to fill the first three places. Therefore, the number of arrangements of 3 different objects from 10 distinct objects is given by:
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{10}P_{3} \ = \ {\large \frac{10!}{(10 \ – \ 3)!} } }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{10!}{7!} } }\)
\(\\[16pt]\hspace{4em} \require{cancel} = \ {\small {\large \frac{10 \ \times \ 9 \ \times \ 8 \ \times {\cancel{7!}}}{{\cancel{7!}}} } }\)
\(\\[12pt]\hspace{4em} = \ {\small 720 }\)
There are 720 ways the first three places can be filled.
\(\\[1pt]\)
\({\small 5.\enspace}\) By writing out all the possible arrangements of \({\small {D}_{1}{E}_{1}{E}_{2}{D}_{2} }\), show that there are \( \frac{4!}{2!(2)!} = {\small 6 \ }\) different arrangements of the letters of the word
DEED.
The six different arangements are:
\(\\[1pt]\)
\(\\[7pt]{\small \hspace{1.2em} \ – \ DDEE }\)
\(\\[7pt]{\small \hspace{1.2em} \ – \ DEDE }\)
\(\\[7pt]{\small \hspace{1.2em} \ – \ DEED }\)
\(\\[7pt]{\small \hspace{1.2em} \ – \ EDDE }\)
\(\\[7pt]{\small \hspace{1.2em} \ – \ EDED }\)
\({\small \hspace{1.2em} \ – \ EEDD }\)
\(\\[1pt]\)
\({\small 6.\enspace}\) A typist has five letters and five addressed envelopes. In how many different ways can the letters be placed in each envelope without getting every letter in the right envelope? If the letters are placed in the envelopes at random what is the probability that each letter is in its correct envelope?
The number of possible arrangements for 5 letters to be placed in 5 envelopes is:
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{5}P_{5} \ = \ {\large \frac{5!}{(5 \ – \ 5)!} } }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{5!}{0!} } }\)
\(\\[10pt]\hspace{4em}{\small = \ 5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1}\)
\(\\[10pt]\hspace{4em}{\small = \ 120 }\)
There is only 1 possible arrangement for each letter to be placed in the right envelope. Therefore, there are \({\small 120 \ – \ 1 \ = \ 119}\) ways for the letters to be placed in each envelope without getting every letter in the right envelope.
\(\\[1pt]\)
The probability for each letter is in its correct envelope is \({\small \frac{1}{120}}\).
\(\\[1pt]\)
\({\small 7.\enspace}\) How many different arrangements can be made of the letters in the word
STATISTICS?
\(\\[7pt]{\small \hspace{1.2em} STATISTICS }\)
\(\\[7pt]{\small \hspace{1.2em} S \ = \ }\) 3 letter
\(\\[7pt]{\small \hspace{1.2em} T \ = \ }\) 3 letter
\(\\[7pt]{\small \hspace{1.2em} I \ = \ }\) 2 letter
\(\\[7pt]{\small \hspace{1.2em} A \ = \ }\) 1 letter
\(\\[10pt]{\small \hspace{1.2em} C \ = \ }\) 1 letter
There are 10 letters in total, with 3 identical letters (S, T and I). The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{10!}{3! \ \times \ 3! \ \times \ 2!} } }\)
\(\\[12pt]\hspace{4em} = \ {\small 50 \ 400 }\)
There are 50400 ways the first three places can be filled.
\(\\[1pt]\)
\({\small 8.\enspace \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) Calculate the number of arrangements of the letters in the word
NUMBER.
\(\\[1pt]\)
\({\small \hspace{2.3em}\textrm{(b)}.\hspace{0.8em}}\) How many of the arrrangements in part (a) begin and end with a vowel?
\(\\[10pt]{\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em} NUMBER }\)
There are 6 distinct letters without any identical letters.
\(\\[1pt]\)
The number of arrangements of 6 letters taken from 6 distinct letters is,
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{6}P_{6} \ = \ {\large \frac{6!}{(6 \ – \ 6)!} } }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{6!}{0!} } }\)
\(\\[10pt]\hspace{4em}{\small = \ 6 \ \times \ 5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1}\)
\(\\[10pt]\hspace{4em}{\small = \ 720 }\)
There are 720 ways to arrange the letters in the word NUMBER.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) The possible arrangements that begin and end with a vowel are:
\(\\[1pt]\)
\(\\[10pt]{\small \hspace{1.2em} U \ \_ \ \_ \ \_ \ \_ \ E \ \ \textrm{or} \ \ E \ \_ \ \_ \ \_ \ \_ \ U }\)
There are 2 possible vowel arrangements, U and E or E and U. And also, in between the 2 vowels, there are 4 possible letters to be arranged from 4 distinct letters, thus:
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{\small = \ ^{2}P_{2} \ \times \ ^{4}P_{4} }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{2!}{(2 \ – \ 2)!} \ \times \ \frac{4!}{(4 \ – \ 4)!} } }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{2!}{0!} \ \times \ \frac{4!}{0!} } }\)
\(\\[10pt]\hspace{4em}{\small = \ 2! \ \times \ 4!}\)
\(\\[10pt]\hspace{4em}{\small = \ 48 }\)
There are 48 ways to arrange the letters in the word NUMBER that begin and end with a vowel.
\(\\[1pt]\)
\({\small 9.\enspace}\) How many different numbers can be formed by taking one, two, three and four digits from the digits 1, 2, 7 and 8, if repetitions are not allowed?
One of these numbers is chosen at random. What is the probability that it is greater than 200?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) There are 4 distinct digits: \( 1, 2, 7 \ \textrm{and} \ \ 8 \).
\(\\[1pt]\)
For one-digit number, the number of possible arrangements is \( \ {\small ^{4}P_{1} }\).
For two-digit number, it is \( \ {\small ^{4}P_{2} }\).
For three-digit number, it is \( \ {\small ^{4}P_{3} }\).
\(\\[10pt]\)For four-digit number, it is \( \ {\small ^{4}P_{4} }\).
The total number of possible arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{0.5em}{\small = \ ^{4}P_{1} \ + \ ^{4}P_{2} \ + \ ^{4}P_{3} \ + \ ^{4}P_{4} }\)
\(\\[15pt]\hspace{0.5em}{\small = \ {\large \frac{4!}{(4 \ – \ 1)!} + \frac{4!}{(4 \ – \ 2)!} + \frac{4!}{(4 \ – \ 3)!} + \frac{4!}{(4 \ – \ 4)!} } }\)
\(\\[10pt]\hspace{0.5em}{\small = \ 4 \ + \ 12 \ + \ 24 \ + \ 24 }\)
\(\\[10pt]\hspace{0.5em}{\small = \ 64 }\)
There are 64 ways to form different numbers from the digits.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) The probability of one chosen number is greater than 200 is the sum of the number of arrangements of three-digit number greater than 200 and the number of arrangements of four-digit number divided by the total number of possible arrangements.
\(\\[1pt]\)
For the three-digit number greater than 200,
– the first digit can only be occupied by the digits 2, 7 and 8. Hence, there are \( \ {\small ^{3}P_{1} \ = \ }\) 3 possible arrangements.
– the second and last digit can be filled by the remaining three digits that have not been used in the first digit. Therefore, there are 2 digits to be arranged from 3 distinct digits.
\(\\[1pt]\)
The number or arrangements for the three-digit number greater than 200 is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{\small = \ 3 \ \times \ ^{3}P_{2} }\)
\(\\[15pt]\hspace{4em}{ = \ 3 {\large \ \times \ \frac{3!}{(3 \ – \ 2)!} } }\)
\(\\[10pt]\hspace{4em}{\small = \ 3 \ \times \ 3!}\)
\(\\[12pt]\hspace{4em}{\small = \ 18 }\)
Meanwhile, for the four-digit number, there are \( \ {\small ^{4}P_{4} \ = \ }\) 24 possible arrangements.
\(\\[1pt]\)
Thus, the probability is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{18 \ + \ 24}{64} } }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{42}{64} } }\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{21}{32} } }\)
\(\\[1pt]\)
EXERCISE 5B:
\({\small 1.\enspace}\) How many three-card hands can be dealt from a pack of 52 cards?
There are 3 cards to choose from 52 cards, the number of choices is,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{52}{3} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 52 \ \times \ 51 \ \times \ 50 }{ 3 \ \times \ 2 \ \times \ 1 } } \)
\(\hspace{3em} \ = \ 22 \ 100 \)
\(\\[1pt]\)
\({\small 2.\enspace}\) From a group of 30 boys and 32 girls, two girls and two boys are to be chosen to represent their school. How many possible selections are there?
There are 2 boys to choose from 30 boys. The number of possible selections for boys is \({\small \ {\displaystyle \binom{30}{2} } }\).
\(\\[1pt]\)
There are 2 girls to choose from 32 girls. The number of possible selections for girls is \({\small \ {\displaystyle \binom{32}{2} } }\).
\(\\[1pt]\)
The total number of possible selections for boys and girls is,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{30}{2} \ \times \ \binom{32}{2} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 30 \ \times \ 29 }{ 2 \ \times \ 1 } } \ \times \ {\large \frac{ 32 \ \times \ 31 }{ 2 \ \times \ 1 } } \)
\(\\[10pt]\hspace{3em} \ = \ 435 \ \times \ 496 \)
\(\hspace{3em} \ = \ 215 \ 760 \)
\(\\[1pt]\)
\({\small 3.\enspace}\) A history exam paper contains eight questions, four in Part A and four in Part B. Candidates are required to attempt five questions. In how many ways can this be done if
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) there are no restrictions,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) at least two questions from Part A and at least two questions from Part B must be attempted?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) When there are no restrictions, there are 5 questions to be attempted from the required 8 questions. The number of possible selections is,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{8}{5} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 8 \ \times \ 7 \ \times \ 6 \ \times \ 5 \ \times \ 4 }{ 5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \)
\(\hspace{3em} \ = \ 56 \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) To answer five questions with at least two questions from Part A and at least two questions from Part B must be attempted, there are 2 possibilities for the candidates to fulfill this requirement:
\(\\[1pt]\)
\(\\[10pt]\hspace{1.2em}\) They can either:
\(\hspace{2em}\) 1.\(\hspace{0.8em}\) answer 2 questions from Part A and 3 questions from Part B, or
\(\\[1pt]\)
\(\hspace{2em}\) 2.\(\hspace{0.8em}\) answer 3 questions from Part A and 2 questions from Part B.
\(\\[1pt]\)
The number of possible selections for the first possibility is,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{4}{2} \ \times \ \binom{4}{3} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 4 \ \times \ 3 }{ 2 \ \times \ 1 } } \ \times \ {\large \frac{ 4 \ \times \ 3 \ \times \ 2 }{ 3 \ \times \ 2 \ \times \ 1 } } \)
\(\\[10pt]\hspace{3em} \ = \ 6 \ \times \ 4 \)
\(\hspace{3em} \ = \ 24 \)
\(\\[1pt]\)
The number of possible selections for the second possibility is,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{4}{3} \ \times \ \binom{4}{2} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 4 \ \times \ 3 \ \times \ 2 }{ 3 \ \times \ 2 \ \times \ 1 } \ \times \ {\large \frac{ 4 \ \times \ 3 }{ 2 \ \times \ 1 } } } \)
\(\\[10pt]\hspace{3em} \ = \ 4 \ \times \ 6 \)
\(\hspace{3em} \ = \ 24 \)
\(\\[1pt]\)
The total number of possible selections is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} \ = \ 24 \ + \ 24 \)
\(\hspace{3em} \ = \ 48 \)
\(\\[1pt]\)
\({\small 4.\enspace}\) A committee of three people is to be selected from four women and five men. The rules state that there must be at least one man and one woman on the committee. In how many different ways can the committee be chosen?
Subsequently one of the men and one of the women marry each other. The rules also state that a married couple may not both serve on the committee. In how many ways can the committee be chosen now?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) Similar to how we tackle question 3 part (b), there are 2 possibilities the committee can be choosen:
\(\\[1pt]\)
\(\\[10pt]\hspace{1.2em}\) They can either:
\(\hspace{2em}\) 1.\(\hspace{0.8em}\) choose one man and two women, or
\(\\[1pt]\)
\(\hspace{2em}\) 2.\(\hspace{0.8em}\) choose two men and one woman.
\(\\[1pt]\)
The number of possible selections for the first possibility is,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{5}{1} \ \times \ \binom{4}{2} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 5 }{ 1 } \ \times \ {\large \frac{ 4 \ \times \ 3 }{ 2 \ \times \ 1 } } } \)
\(\\[10pt]\hspace{3em} \ = \ 5 \ \times \ 6 \)
\(\hspace{3em} \ = \ 30 \)
\(\\[1pt]\)
The number of possible selections for the second possibility is,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{5}{2} \ \times \ \binom{4}{1} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 5 \ \times \ 4 }{ 2 \ \times \ 1 } \ \times \ {\large \frac{ 4 }{ 1 } } } \)
\(\\[10pt]\hspace{3em} \ = \ 10 \ \times \ 4 \)
\(\hspace{3em} \ = \ 40 \)
\(\\[1pt]\)
The total number of possible selections is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} \ = \ 30 \ + \ 40 \)
\(\hspace{3em} \ = \ 70 \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) We can find the number of ways for the committee to be chosen by subtracting the total number of possible selections from the number of possible selections with the married couple in the committee.
\(\\[1pt]\)
So, let us consider when the married couple is in the committee.
\(\\[1pt]\)
In the case of the committee consists of 1 man and 2 women, the slot for 1 man and 1 woman are already occupied by the married couple.
\(\\[1pt]\)
Therefore, we are left with only 1 more slot for 1 woman. To choose 1 woman from the remaining 3 women,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{3}{1} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 3 }{ 1 } } \)
\(\hspace{3em} \ = \ 3 \)
\(\\[1pt]\)
Meanwhile, when the committee consists of 2 men and 1 woman, the slot for 1 man and 1 woman are already occupied by the married couple.
\(\\[1pt]\)
Therefore, we are left with only 1 more slot for 1 man. To choose 1 man from the remaining 4 men,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{4}{1} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 4 }{ 1 } } \)
\(\hspace{3em} \ = \ 4 \)
\(\\[1pt]\)
The number of possible selections with the married couple in the committee is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} \ = \ 3 \ + \ 4 \)
\(\hspace{3em} \ = \ 7 \)
\(\\[1pt]\)
The solution for the number of ways for the committee to be chosen with no married couple serve together is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} \ = \ 70 \ – \ 7 \)
\(\hspace{3em} \ = \ 63 \)
\(\\[1pt]\)
\({\small 5.\enspace}\) A box of one dozen eggs contains one that is bad. If three eggs are chosen at random, what is the probability that one of them will be bad?
The probability is the number of possible selections of selecting 3 eggs from a dozen eggs with 1 bad egg divided by the the number of possible selections of selecting 3 eggs from a dozen eggs.
\(\\[1pt]\)
For the numerator part of the probability,
\(\\[1pt]\)
There are 11 good eggs and 1 bad egg. To choose 3 eggs that contain 2 good eggs and 1 bad egg,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{11}{2} \ \times \ \binom{1}{1} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 11 \ \times \ 10 }{ 2 \ \times \ 1 } \ \times \ 1 } \)
\(\\[10pt]\hspace{3em} \ = \ 55 \ \times \ 1 \)
\(\\[15pt]\hspace{3em} \ = \ 55 \)
For the denominator part of the probability,
\(\\[1pt]\)
There are 12 eggs. To choose 3 eggs,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{12}{3} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 12 \ \times \ 11 \ \times \ 10 }{ 3 \ \times \ 2 \ \times \ 1 } } \)
\(\\[15pt]\hspace{3em} \ = \ 220 \)
Hence, the probability is,
\(\\[1pt]\)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 55 }{ 220 } } \)
\(\\[10pt]\hspace{3em} \ = \ {\large \frac{ 1 }{ 4 } } \)
\(\\[1pt]\)
\({\small 6.\enspace}\) In a game of bridge the pack of 52 cards is shared equally between all four players. What is the probability that one particular player has no hearts?
With 52 cards equally shared between 4 players, each of the player will get 13 cards.
\(\\[1pt]\)
Also, there are 13 ‘hearts’ cards and 39 ‘non-hearts’ cards in the deck.
\(\\[1pt]\)
The probability for a particular player to be dealt with ‘non-hearts’ cards is,
\(\\[1pt]\)
\(\\[35pt]\hspace{1.8em} \ = \ \frac{{\displaystyle \binom{39}{13} }}{ {\displaystyle \binom{52}{13} } } \)
\(\\[20pt]\hspace{1.8em} \ = \ {\large \frac{ 39! }{ 13! \ \times \ (39 \ – \ 13)! } \div {\large \frac{ 52! }{ 13! \ \times \ (52 \ – \ 13)! } } } \)
\(\hspace{1.8em} \ \approx \ 0.0128 \)
\(\\[1pt]\)
\({\small 7.\enspace}\) A bag contains 20 chocolates, 15 toffees and 12 peppermints. If three sweets are chosen at random, what is the probability that they are
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) all different,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) all chocolates,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) all the same,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}\) all not chocolates?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) There are 3 sweets to be chosen from 47 sweets. To choose all different sweets, 1 chocolate, 1 toffee and 1 peppermint; the probability is,
\(\\[1pt]\)
\(\\[35pt]\hspace{1.2em} \ = \ \frac{{\displaystyle \binom{20}{1} \times \binom{15}{1} \times \binom{12}{1} }}{ {\displaystyle \binom{47}{3} } } \)
\(\\[20pt]\hspace{1.2em} \ = \ {\large \frac{ 20 \ \times \ 15 \ \times \ 12 }{ 16 \ 215} } \)
\(\hspace{1.2em} \ \approx \ 0.222 \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) There are 3 sweets to be chosen from 47 sweets. To choose all chocolates, 3 chocolates; the probability is,
\(\\[1pt]\)
\(\\[35pt]\hspace{1.2em} \ = \ \frac{{\displaystyle \binom{20}{3} }}{ {\displaystyle \binom{47}{3} } } \)
\(\\[20pt]\hspace{1.2em} \ = \ {\large \frac{ 1 \ 140 }{ 16 \ 215} } \)
\(\hspace{1.2em} \ \approx \ 0.070 \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) There are 3 sweets to be chosen from 47 sweets. To choose all the same sweets, 3 chocolates or 3 toffees or 3 peppermints; the probability is,
\(\\[1pt]\)
\(\\[35pt]\hspace{1.2em} \ = \ \frac{{\displaystyle \binom{20}{3} \ + \ \binom{15}{3} \ + \ \binom{12}{3} }}{ {\displaystyle \binom{47}{3} } } \)
\(\\[20pt]\hspace{1.2em} \ = \ {\large \frac{ 1 \ 140 \ + \ 455 \ + \ 220 }{ 16 \ 215} } \)
\(\hspace{1.2em} \ \approx \ 0.112 \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}\) There are 3 sweets to be chosen from 47 sweets. To choose all not chocolates, 3 ‘non-chocolates’ from 27 ‘non-chocolates’ sweets; the probability is,
\(\\[1pt]\)
\(\\[35pt]\hspace{1.2em} \ = \ \frac{{\displaystyle \binom{27}{3} }}{ {\displaystyle \binom{47}{3} } } \)
\(\\[20pt]\hspace{1.2em} \ = \ {\large \frac{ 2 \ 925 }{ 16 \ 215} } \)
\(\hspace{1.2em} \ \approx \ 0.180 \)
\(\\[1pt]\)
\({\small 8.\enspace}\) Show that \( \displaystyle \binom{n}{r} \ = \ \binom{n}{n \ – \ r} \).
By definition,
\(\\[1pt]\)
\( \displaystyle \binom{n}{r} \ = \ \frac{n!}{r!(n-r)!} \)
\(\\[1pt]\)
\(\\[15pt]\)Then,
\(\\[23pt] \displaystyle \binom{n}{n-r} \ = \ \frac{n!}{(n-r)![n-(n-r)]!} \)
\( \\[23pt]\hspace{3.8em} \ = \ \displaystyle \frac{n!}{(n-r)!r!} \)
They are both equal.
\(\\[1pt]\)
\({\small 9.\enspace}\) Show that the number of permutations of
n objects of which
r are of one kind and
n – r are of another kind is \( \displaystyle \binom{n}{r} \).
The number of permutations of n objects with r identical objects and n – r identical objects is,
\(\\[1pt]\)
\(\\[30pt]\hspace{3em} \displaystyle = \ \frac{n!}{r!(n-r)!} \)
\(\\[20pt]\hspace{3em} \displaystyle = \ \binom{n}{r} \)
\(\\[1pt]\)
EXERCISE 5C:
\({\small 1.\enspace}\) The letters of the word
CONSTANTINOPLE are written on 14 cards, one on each card. The cards are shuffled and then arranged in a straight line.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many different possible arrangements are there?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many arrangements begin with
P?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) How many arrangements start and end with
O?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}\) How many arrangements are there where no two vowels are next to each other?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) \(\\[7pt]{\small CONSTANTINOPLE }\)
\(\\[7pt]{\small \hspace{3em} N \ = \ }\) 3 letter
\(\\[7pt]{\small \hspace{3em} O \ = \ }\) 2 letter
\(\\[7pt]{\small \hspace{3em} T \ = \ }\) 2 letter
\(\\[7pt]{\small \hspace{3em} C \ = \ }\) 1 letter
\(\\[7pt]{\small \hspace{3em} S \ = \ }\) 1 letter
\(\\[7pt]{\small \hspace{3em} P \ = \ }\) 1 letter
\(\\[7pt]{\small \hspace{3em} L \ = \ }\) 1 letter
\(\\[7pt]{\small \hspace{3em} A \ = \ }\) 1 letter
\(\\[7pt]{\small \hspace{3em} I \ = \ }\) 1 letter
\(\\[10pt]{\small \hspace{3em} E \ = \ }\) 1 letter
There are 14 letters in total, with 3
identical letters (N, O and T). The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{14!}{3! \ \times \ 2! \ \times \ 2!} } }\)
\(\\[20pt]\hspace{4em} = \ {\small 3 \ 632 \ 428 \ 800 }\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) With P always as the first letter in our arrangements, we now have 13 available letters in total, with 3
identical letters (N, O and T). The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{13!}{3! \ \times \ 2! \ \times \ 2!} } }\)
\(\\[20pt]\hspace{4em} = \ {\small 259 \ 459 \ 200 }\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) With O always as the first and the last letter in our arrangements, we now have 12 available letters in total, with 2
identical letters (N and T). The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{12!}{ 3! \ \times \ 2!} } }\)
\(\\[20pt]\hspace{4em} = \ {\small 39 \ 916 \ 800 }\)
\({\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}\) We will approach this problem by using the analogy of boxes and spaces between the boxes.
\(\\[1pt]\)
There will be
n boxes and
n + 1 spaces.
\(\\[1pt]\)
Since there are no two vowels next to each other, we will separate the vowels and the consonants. There are
9 consonants and 5 vowels.
\(\\[1pt]\)
The consonants will fill in the
9 boxes while the vowels will fill in
10 spaces between the boxes.
\(\\[1pt]\)
To have a better picture of the arrangements, you can take a look at the illustration below.
\(\\[1pt]\)
\(\\[1pt]\)
For the consonants, we have 9 letters with 2
identical letters (N and T). The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{9!}{ 3! \ \times \ 2!} } }\)
\(\\[20pt]\hspace{4em} = \ {\small 30 \ 240 }\)
For the vowels, we have 5 letters with 1
identical letter (O) to be filled in 10 spaces. The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{^{10}P_{5}}{ 2!} } }\)
\(\\[20pt]\hspace{4em} = \ {\small 15 \ 120 }\)
The total number of possible arrangements is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} \ = \ 30 \ 240 \ \times \ 15 \ 120 \)
\(\hspace{3em} \ = \ 457 \ 228 \ 800 \)
\(\\[1pt]\)
\({\small 2.\enspace}\) A coin is tossed 10 times.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many different sequences of heads and tails are possible?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many different sequences containing six heads and four tails are possible?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) What is the probability of getting six heads and four tails?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) There are \({\small {2}^{n} }\) sequences with n number of tosses. For 10 tosses, there are \({\small {2}^{10} \ = \ 1024 }\) different sequences of heads and tails.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) To select 6 heads out of 10 tosses or 4 tails out of 10 tosses, the number of possible selections is,
\(\\[1pt]\)
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{10}{6} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 \ \times \ 7 \ \times \ 6 \ \times \ 5 }{ 6 \ \times \ 5 \ \times \ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \)
\(\hspace{3em} \ = \ 210 \)
\(\\[12pt]\) Or,
\(\\[20pt]\hspace{3em} \ = \ {\displaystyle \binom{10}{4} } \)
\(\\[15pt]\hspace{3em} \ = \ {\large \frac{ 10 \ \times \ 9 \ \times \ 8 \ \times \ 7 }{ 4 \ \times \ 3 \ \times \ 2 \ \times \ 1 } } \)
\(\hspace{3em} \ = \ 210 \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) By combining the result in part (a) and (b), the probability is,
\(\\[1pt]\)
\(\\[25pt]\hspace{1.2em} \ = \ \frac{{\displaystyle \binom{10}{6} }}{\LARGE{{2}^{10}} } \)
\(\\[20pt]\hspace{1.2em} \ = \ {\large \frac{ 210 }{ 1024} } \)
\(\hspace{1.2em} \ \approx \ 0.205 \)
\(\\[1pt]\)
\({\small 3.\enspace}\) Eight cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture cards, 20 odd cards and 20 even cards.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many different sequences of eight cards are possible?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many of the sequences in part (a) will contain three picture cards, three odd-numbered cards and two even-numbered cards?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) Use parts (a) and (b) to determine the probability of getting three picture cards, three odd-numbered cards and two even-numbered cards if eight cards are selected with replacement from a standard pack of 52 playing cards.
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) There are \({\small {52}^{n} }\) sequences with n number of selected cards. For 8 selected cards, there are \({\small {52}^{8} \ = \ 5.346 \times {10}^{13} }\) possible sequences.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) For the picture cards, we select 3 picture cards out of 8 cards. Also, there are 12 picture cards available for the selection.
\(\\[1pt]\)
The number of sequences of 3 picture cards from 8 cards with replacement is,
\(\\[1pt]\)
\(\\[20pt]\hspace{1.2em} \ = \ {\displaystyle \binom{8}{3} } \ \times \ {12}^{3} \)
\(\\[20pt]\hspace{1.2em} \ = \ 96 \ 768 \)
For the odd-numbered cards, we select 3 odd-numbered cards out of 5 remaining cards (3 of 8 cards have been used by the picture cards). Also, there are 20 odd-numbered cards available for the selection.
\(\\[1pt]\)
The number of sequences of 3 odd-numbered cards from 5 cards with replacement is,
\(\\[1pt]\)
\(\\[20pt]\hspace{1.2em} \ = \ {\displaystyle \binom{5}{3} } \ \times \ {20}^{3} \)
\(\\[20pt]\hspace{1.2em} \ = \ 80 \ 000 \)
For the even-numbered cards, we select 2 even-numbered cards out of 2 remaining cards (6 of 8 cards have been used by the picture and odd-numbered cards). Also, there are 20 even-numbered cards available for the selection.
\(\\[1pt]\)
The number of sequences of 2 odd-numbered cards from 2 cards with replacement is,
\(\\[1pt]\)
\(\\[20pt]\hspace{1.2em} \ = \ {\displaystyle \binom{2}{2} } \ \times \ {20}^{2} \)
\(\\[20pt]\hspace{1.2em} \ = \ 400 \)
The total number of sequences is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} \ = \ 96 \ 768 \ \times \ 80 \ 000 \ \times \ 400 \)
\(\hspace{3em} \ = \ 3.097 \times {10}^{12} \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) By combining the result in part (a) and (b), the probability is,
\(\\[1pt]\)
\(\\[25pt]\hspace{1.2em} \ = \ {\large \frac{ 3.097 \times {10}^{12} }{ 5.346 \times {10}^{13} } } \)
\(\hspace{1.2em} \ \approx \ 0.0579 \)
\(\\[1pt]\)
\({\small 4.\enspace}\) Eight women and five men are standing in a line.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many arrangements are possible if any individual can stand in any position?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) In how many arrangements will all five men be standing next to one another?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) In how many arrangements will no two men be standing next to one another?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) There are 13 people available for the arrangements. The number of possible arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{13}P_{13} \ = \ {\large \frac{13!}{(13 \ – \ 13)!} } }\)
\(\\[15pt]\hspace{4.4em}{ = \ {\large \frac{13!}{0!} } }\)
\(\\[20pt]\hspace{4.4em}{\small = \ 6 \ 227 \ 020 \ 800 }\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) Consider the five men as a single unit. We will now have 9 units or slots to fill in, 8 women and a block of 5 men.
\(\\[1pt]\)
There are \( {\small ^{9}P_{9} }\) arrangements for these units. Also, within the block of 5 men, there are \( {\small ^{5}P_{5} }\) arrangements can be made.
\(\\[1pt]\)
Therefore, the number of possible arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{3.6em} \ = \ {\small ^{9}P_{9} \ \times \ ^{5}P_{5} }\)
\(\\[15pt]\hspace{4em} {\small = } \ {\large \frac{9!}{(9 \ – \ 9)!} \ \times \ \frac{5!}{(5 \ – \ 5)!} } \)
\(\\[15pt]\hspace{4em}{ {\small = } \ {\large \frac{9!}{0!} \ \times \ \frac{5!}{0!} } }\)
\(\\[20pt]\hspace{4em}{\small = \ 43 \ 545 \ 600 }\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) Similar as Question 1 part (d), we will approach this problem by using the analogy of boxes and spaces between the boxes.
\(\\[1pt]\)
There will be n boxes and n + 1 spaces.
\(\\[1pt]\)
Since there are no two men standing next to each other, we will separate the men and women. There are 8 women and 5 men.
\(\\[1pt]\)
The women will fill in the 8 boxes while the men will fill in 9 spaces between the boxes.
\(\\[1pt]\)
For the women, the number of arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{8}P_{8} \ = \ {\large \frac{8!}{(8 \ – \ 8)!} } }\)
\(\\[15pt]\hspace{4em}{ {\small = } \ {\large \frac{8!}{0!} } }\)
\(\\[20pt]\hspace{4em}{\small = \ 40 \ 320 }\)
For the men, we have 5 men to fill in 9 spaces. The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{9}P_{5} \ = \ {\large \frac{9!}{(9 \ – \ 5)!} } }\)
\(\\[15pt]\hspace{4em}{ {\small = } \ {\large \frac{9!}{4!} } }\)
\(\\[20pt]\hspace{4em}{\small = \ 15 \ 120 }\)
The total number of possible arrangements is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} {\small \ = \ 40 \ 320 \ \times \ 15 \ 120 }\)
\(\hspace{3em} {\small \ = \ 609 \ 638 \ 400 }\)
\(\\[1pt]\)
\({\small 5.\enspace}\) Each of the digits 1, 1, 2, 3, 3, 4, 6 is written on a separate card. The seven cards are then laid out in a row to form a 7-digit number.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many distinct 7-digit numbers are there?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many of these 7-digit numbers are even?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) How many of these 7-digit numbers are divisible by 4?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}\) How many of these 7-digit numbers start and end with the same digit?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) There are 7 digits in total, with 2 identical digits (1 and 3). The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{7!}{ 2! \ \times \ 2!} } }\)
\(\\[20pt]\hspace{4em} = \ {\small 1 \ 260 }\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) The 7-digit numbers are even if their last digit is an even number.
\(\\[1pt]\)
There are 3 possible digits for the last digit of the 7-digit numbers: 2, 4 and 6.
\(\\[1pt]\)
For the first 6 digits of the 7-digit numbers, the number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{6!}{ 2! \ \times \ 2!} } }\)
\(\\[20pt]\hspace{4em} = \ {\small 180 }\)
Then, the total number of permutations is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} {\small \ = \ 3 \ \times \ 180 }\)
\(\\[20pt]\hspace{3em} {\small \ = \ 540 }\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) The 7-digit numbers are divisible by 4 if their last two-digits are divisible by 4.
\(\\[1pt]\)
Let’s evaluate the possible choices for the last two-digits that are divisible by 4.
\(\\[1pt]\)
The possible choices for the last two-digits are: 12, 16, 24, 32, 36 and 64.
\(\\[1pt]\)
For the last two-digits, 24 and 64, there are remaining 5 digits in total, with 2 identical digits (1 and 3). The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{5!}{ 2! \ \times \ 2!} } }\)
\(\\[20pt]\hspace{4em} = \ {\small 30 }\)
For the last two-digits, 12 and 16, there are remaining 5 digits in total, with 1 identical digit (3). The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{5!}{ 2! } } }\)
\(\\[20pt]\hspace{4em} = \ {\small 60 }\)
For the last two-digits, 32 and 36, there are remaining 5 digits in total, with 1 identical digit (1). The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{5!}{ 2! } } }\)
\(\\[20pt]\hspace{4em} = \ {\small 60 }\)
The total number of possible arrangements is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} {\small \ = \ 30 \times 2 \ + \ 60 \times \ 4 }\)
\(\\[20pt]\hspace{3em} {\small \ = \ 300 }\)
\({\small \hspace{1.2em}\textrm{(d)}.\hspace{0.8em}}\) To start and end with the same digit, only 2 possible digits can satisfy the requirement, digit 1 and digit 3.
\(\\[1pt]\)
If the start and end digit is 1, there are 5 remaining digits in between with an identical digit 3.
\(\\[1pt]\)
If the start and end digit is 3, there are 5 remaining digits in between with an identical digit 1.
\(\\[1pt]\)
The total number of possible arrangements is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} {\small \ = \ 2 \times {\large \frac{5!}{ 2! } } }\)
\(\hspace{3em} {\small \ = \ 120 }\)
\(\\[1pt]\)
\({\small 6.\enspace}\) Three families, the Mehtas, the Mupondas and the Lams, go to the cinema together to watch a film. Mr and Mrs Mehta take their daughter Indira, Mr and Mrs Muponda take their sons Paul and John, and Mrs Lam takes her children Susi, Kim and Lee. The families occupy a single row with eleven seats.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) In how many ways could the eleven people be seated if there were no restriction?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) In how many ways could the eleven people sit down so that the members of each family are all sitting together?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) In how many of the arrangements will no two adults be sitting next to one another?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) There are 11 people available for the arrangements. The number of possible arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{11}P_{11} \ = \ {\large \frac{11!}{(11 \ – \ 11)!} } }\)
\(\\[15pt]\hspace{4.4em}{ = \ {\large \frac{11!}{0!} } }\)
\(\\[20pt]\hspace{4.4em}{\small = \ 39 \ 916 \ 800 }\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) Each of the families can be grouped as a single unit. There are 3 units: the Mehtas, the Mupondas and the Lams.
\(\\[1pt]\)
With the Mehtas consists of 3 people, there are \( {\small ^{3}P_{3} }\) ways of arrangements.
\(\\[1pt]\)
With the Mupondas consists of 4 people, there are \( {\small ^{4}P_{4} }\) ways of arrangements.
\(\\[1pt]\)
With the Lams consists of 4 people, there are \( {\small ^{4}P_{4} }\) ways of arrangements.
\(\\[1pt]\)
Therefore, the number of possible arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{3.6em} \ = \ {\small ^{3}P_{3} \times ^{3}P_{3} \times ^{4}P_{4} \times ^{4}P_{4} }\)
\(\\[15pt]\hspace{3.6em} \ = \ {\small 6 \times 6 \times 24 \times 24 }\)
\(\\[20pt]\hspace{4em}{\small = \ 20 \ 736 }\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em}}\) Since there will be no two adults sitting next to each other, we will separate the adults and children. There are 5 adults and 6 children.
\(\\[1pt]\)
The children will fill in the 6 boxes while the adults will fill in 7 spaces between the boxes.
\(\\[1pt]\)
For the children, the number of arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{6}P_{6} \ = \ {\large \frac{6!}{(6 \ – \ 6)!} } }\)
\(\\[15pt]\hspace{4em}{ {\small = } \ {\large \frac{6!}{0!} } }\)
\(\\[20pt]\hspace{4em}{\small = \ 720 }\)
For the adults, we have 5 adults to fill in 7 spaces. The number of permutations is,
\(\\[1pt]\)
\(\\[15pt]\hspace{2.2em}{\small ^{7}P_{5} \ = \ {\large \frac{7!}{(7 \ – \ 5)!} } }\)
\(\\[15pt]\hspace{4em}{ {\small = } \ {\large \frac{7!}{2!} } }\)
\(\\[20pt]\hspace{4em}{\small = \ 2 \ 520 }\)
The total number of possible arrangements is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} {\small \ = \ 720 \ \times \ 2 \ 520 }\)
\(\hspace{3em} {\small \ = \ 1 \ 814 \ 400 }\)
\(\\[1pt]\)
\({\small 7.\enspace}\) The letters of the word
POSSESSES are written on nine cards, one on each card. The cards are shuffled and four of them are selected and arranged in a straight line.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many possible selections are there of four letters?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) How many arrangements are there of four letters?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) There are 9 letters in total, with 2 identical letters, (S and E).
\(\\[1pt]\)
We can split the possible selections into different cases depending on the number of Ss chosen.
\(\\[1pt]\)
\(\\[10pt]\hspace{1.2em}\) Case 1: 4 S
The possible selection is: 1 (SSSS).
\(\\[1pt]\)
\(\\[10pt]\hspace{1.2em}\) Case 2: 3 S
The possible selections is: 3 (SSSE, SSSO, SSSP).
\(\\[1pt]\)
\(\\[10pt]\hspace{1.2em}\) Case 3: 2 S
The possible selection is: 4 (SSEE, SSEO, SSEP, SSOP).
\(\\[1pt]\)
\(\\[10pt]\hspace{1.2em}\) Case 4: 1 S
The possible selection is: 3 (SEEO, SEEP, SEOP).
\(\\[1pt]\)
\(\\[10pt]\hspace{1.2em}\) Case 5: no S
The possible selection is: 1 (PEEO).
\(\\[1pt]\)
The total number of selections is \({\small 1+3+4+3+1 \ = \ 12}\).
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) We can count each possible case in part (a) by taking into account the number of identical letters.
\(\\[1pt]\)
For SSSS, the number of arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{4!}{ 4! } } }\)
\(\\[20pt]\hspace{4em} = \ {\small 1 }\)
For SSSE, SSSO, SSSP, the number of arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ 3 \ \times \ {\large \frac{4!}{ 3! } } }\)
\(\\[20pt]\hspace{4em} = \ {\small 12 }\)
For SSEE, the number of arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{4!}{ 2! \ \times \ 2! } } }\)
\(\\[20pt]\hspace{4em} = \ {\small 6 }\)
For SSEO, SSEP, SSOP, SEEO, SEEP, PEEO, the number of arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ 6 \ \times \ {\large \frac{4!}{ 2! } } }\)
\(\\[20pt]\hspace{4em} = \ {\small 72 }\)
For SEOP, the number of arrangements is,
\(\\[1pt]\)
\(\\[15pt]\hspace{4em}{ = \ {\large \frac{4!}{ 0! } } }\)
\(\\[20pt]\hspace{4em} = \ {\small 24 }\)
The total number of possible arrangements is,
\(\\[1pt]\)
\(\\[10pt]\hspace{3em} {\small \ = 1 \ + \ 12 \ + \ 6 \ + \ 72 \ + \ 24 }\)
\(\hspace{3em} {\small \ = \ 115 }\)
\(\\[1pt]\)
MISCELLANEOUS EXERCISE 5:
\({\small 1.\enspace}\) The judges in a ‘Beautiful Baby’ competition have to arrange 10 babies in order of merit. In how many different ways could this be done? Two babies are to be selected to be photographed. In how many ways can this selection be made?
\(\\[10pt] {\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em} \ 10! }\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em} \ {\displaystyle \binom {10}{2} } }\)
\(\\[1pt]\)
\({\small 2.\enspace}\) In how many ways can a committee of four men and four women be seated in a row if
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) they can sit in any position,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) no one is seated next to a person of the same sex?
\(\\[10pt] {\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em} \ 8! }\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em} \ 2 \times 4! \times 4! }\)
\(\\[1pt]\)
\({\small 3.\enspace}\) How many distinct arrangements are there of the letters in the word
ABRACADABRA?
\(\hspace{4em}{ {\large \frac{11!}{ 5! \ \times \ 2! \ \times \ 2!} } }\)
\(\\[1pt]\)
\({\small 4.\enspace}\) Six people are going to travel in a six-seater minibus but only three of them can drive. In how many different ways can they seat themselves?
\(\hspace{4em}{\small 3 \ \times \ 5! }\)
\(\\[1pt]\)
\({\small 5.\enspace}\) There are eight different books on a bookshelf: three of them are hardbacks and the rest are paperbacks.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) In how many different ways can the books be arranged if all the paperbacks are together and all the hardbacks are together?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) In how many different ways can the books be arranged if all the paperbacks are together?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em} 3! \ \times \ 5! \ \times \ 2! }\)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em} 5! \ \times \ 4!}\)
\(\\[1pt]\)
\({\small 6.\enspace}\) Four boys and two girls sit in a line on stools in front of a coffee bar.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) In how many ways can they arrange themselves so that the two girls are together?
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) In how many ways can they sit if the two girls are not together?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em} 2! \ \times \ 5!}\)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em} 4! \ \times \ ^{5}P_{2}}\)
\(\\[1pt]\)
\({\small 7.\enspace}\) Ten people travel in two cars, a saloon and a Mini. If the saloon has seats for six and the Mini has seats for four, find the number of different ways in which the party can travel, assuming that the order of seating in each car does not matter and all the people can drive.
\(\hspace{4em} {\small {\displaystyle \binom {10}{6} \ } }\) or \( {\small \ {\displaystyle \binom {10}{4} } }\)
\(\\[1pt]\)
\({\small 8.\enspace}\) Giving a brief explanation of your method, calculate the number of different ways in which the letters of the word
TRIANGLES can be arranged if no two vowels may come together.
There are 6 consonants and 3 vowels. The consonants fill in the 6 boxes and the vowels fill in the 7 spaces in between the boxes. The number of arrangements is \({\small 6! \ \times \ ^{7}P_{3} }\)
\(\\[1pt]\)
\({\small 9.\enspace}\) I have seven fruit bars to last the week. Two are apricot, three fig and two peach. I select one bar each day. In how many different orders can I eat the bars?
If I select a fruit bar at random each day, what is the probability that I eat the two appricot ones on consecutive days?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}{\large \frac{7!}{ 3! \ \times \ 2! \ \times \ 2!} } }\)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}{\large \frac{{\small{\displaystyle \binom {2}{2}} }\times {\LARGE \frac{6!}{ 3! \ \times \ 2! }}}{210} } } \)
\(\\[1pt]\)
\({\small 10.\enspace}\) A class contains 30 children, 18 girls and 12 boys. Four complimentary theatre tickets are distributed at random to the children in the class. What is the probability that
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) all four tickets go to the girls,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) two boys and two girls receive tickets?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}{\large \frac{{\small{\displaystyle \binom {18}{4}} }}{{\small{\displaystyle \binom {30}{4}} }} } } \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}{\large \frac{{\small{\displaystyle \binom {12}{2}} } \ \times \ {\small{\displaystyle \binom {18}{2}} }}{{\small{\displaystyle \binom {30}{4}} }} } } \)
\(\\[1pt]\)
\({\small 11.\enspace \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) How many different 7-digit numbers can be formed from the digits 0, 1, 2, 2, 3, 3, 3 assuming that a number cannot start with 0?
\(\\[1pt]\)
\({\small \hspace{2.8em}\textrm{(b)}.\hspace{0.8em}}\) How many of these numbers will end in 0?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}{\large \frac{7!}{ 3! \ \times \ 2! } \ – \ \frac{6!}{ 3! \ \times \ 2! } } }\)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}{\large \frac{6!}{ 3! \ \times \ 2! } } }\)
\(\\[1pt]\)
\({\small 12.\enspace}\) Calculate the number of ways in which three girls and four boys can be seated on a row of seven chairs if each arrangement is to be symmetrical.
To have a symmetrical arrangement, a girl has to be at the centre of the row. Each of the left side and the right side of the centre will have 2 boys and 1 girl.
\(\\[1pt]\)
\(\\[10pt]\) The number of arrangements is,
\({\small \ {\displaystyle \binom{3}{1}} \ \times 4! \ \times 3! }\)
\(\\[1pt]\)
\({\small 13.\enspace}\) Find the number of ways in which
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) 3 people can be arranged in 4 seats,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) 5 people can be arranged in 5 seats.
\(\\[1pt]\)
In a block of 8 seats, 4 are in row A and 4 are in row B. Find the number of ways of arranging 8 people in the 8 seats given that 3 specified people must be in row A.
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em} ^{4}P_{3} }\)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em} ^{5}P_{5} }\)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em} ^{4}P_{3} \ \times \ ^{5}P_{5} }\)
\(\\[1pt]\)
\({\small 14.\enspace}\) Eight different cards, of which four are red and four are black, are dealt to two players so that each receives a hand of four cards. Calculate
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) the total number of different hands which a given player could receive,
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) the probability that each player receives a hand consisting of four cards all of the same colour.
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em} {\displaystyle \binom {8}{4}} } \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em} {\large \frac{{\small{\displaystyle \binom {4}{4}} \ {\large + } \ {\displaystyle \binom {4}{4}}} }{{\small{\displaystyle \binom {8}{4}} }} }}\)
\(\\[1pt]\)
\({\small 15.\enspace}\) A piece of wood of length of 10 cm is to be divided into 3 pieces so that the length of each piece is a whole number of cm, for example, 2 cm, 3 cm and 5 cm.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) List all the different sets of lengths which could be obtained.
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) If one of these sets is selected at random, what is the probability that the lengths of the pieces could be lengths of the sides of a triangle?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) (1, 1, 8), (1, 2, 7), (1, 3, 6), (1, 4, 5), (2, 2, 6), (2, 3, 5), (2, 4, 4), (3, 3, 4)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) Use the triangle inequality: “the sum of the lengths of any two sides must be greater than the third side“.
\(\\[1pt]\)
\({\small \hspace{4em} c \ – \ b \ \lt \ a \ \lt \ c \ + \ b }\)
\(\\[1pt]\)
There are only 2 possible sets: (3, 3, 4) and (2, 4, 4).
\(\\[1pt]\)
The probability is \( \ \frac{2}{8} \ = \ \frac{1}{4} \)
\(\\[1pt]\)
\({\small 16.\enspace}\) Nine persons are to be seated at three tables holding 2, 3 and 4 persons respectively. In how many ways can the groups sitting at the tables be selected, assuming that the order of sitting at the tables does not matter?
\({\small \hspace{4em} {\displaystyle \binom {9}{2} \ \times \ \binom {7}{3} \ \times \ \binom {4}{4} } } \)
\(\\[1pt]\)
\({\small 17.\enspace \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}}\) Calculate the number of different arrangements which can be made using all the letters of the word
BANANA.
\(\\[1pt]\)
\({\small \hspace{2.8em}\textrm{(b)}.\hspace{0.8em}}\) The number of combinations of 2 objects from
n is equal to the number of combinations of 3 objects from
n. Determine
n.
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}{\large \frac{6!}{ 3! \ \times \ 2! } }}\)
\(\\[1pt]\)
\(\\[20pt]{\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em} {\displaystyle \binom {n}{2} \ = \ \binom {n}{3} } }\)
\({\small \hspace{4.9em} n \ = \ 5 }\)
\(\\[1pt]\)
\({\small 18.\enspace}\) A ‘hand’ of 5 cards is dealt from an ordinary pack of 52 playing cards. Show that there are nearly 2.6 million distinct hands and that, of these, 575 757 contain no card from the heart suit.
On three successive occasions a card player is dealt a hand containing no heart. What is the probability of this happening? What conclusion might the player be justifiably reach?
\({\small \hspace{1.2em}\textrm{(a)}.\hspace{0.8em}{\displaystyle \binom {52}{5}} } \)
\(\\[1pt]\)
\({\small \hspace{1.2em}\textrm{(b)}.\hspace{0.8em}}\) There are 52 – 13 = 39 ‘non-hearts’ cards. The number of selections is,
\(\\[1pt]\)
\(\\[25pt]\hspace{3.5em}{\small {\displaystyle \binom {39}{5}} } \)
\({\small \hspace{1.2em}\textrm{(c)}.\hspace{0.8em} {\displaystyle {\binom {37}{5}}}^{3} \ \approx \ 0.0109 }\)
\(\\[1pt]\)
There is something amiss with how the card is dealt/shuffled.
\(\\[1pt]\)
9709 A-Level Past Papers:
\({\small 1.\enspace}\)
9709/51/O/N/21 – Paper 51 November 2021 Probability and Statistics I No 5 \(\\[1pt]\)
Raman and Sanjay are members of a quiz team which has 9 members in total. Two photographs of the quiz team are to be taken.
\(\\[1pt]\)
For the first photograph, the 9 members will stand in a line.
\(\\[1pt]\)
\({\small(\textrm{a}).\hspace{0.8em}}\) How many different arrangements of the 9 members are possible in which Raman will be at the centre of the line?
\(\\[1pt]\)
\({\small(\textrm{b}).\hspace{0.8em}}\) How many different arrangements of the 9 members are possible in which Raman and Sanjay are not next to each other?
\(\\[1pt]\)
For the second photograph, the members will stand in two rows, with 5 in the back row and 4 in the front row.
\(\\[1pt]\)
\({\small(\textrm{c}).\hspace{0.8em}}\) In how many different ways can the 9 members be divided into a group of 5 and a group of 4?
\(\\[1pt]\)
\({\small(\textrm{d}).\hspace{0.8em}}\) For a random division into a group of 5 and a group of 4, find the probability that Raman and
Sanjay are in the same group as each other.
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/CnA6s5-tZdZ/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 2.\enspace}\)
9709/53/M/J/21 – Paper 53 June 2021 Probability and Statistics I No 6 \(\\[1pt]\)
\({\small(\textrm{a}).\hspace{0.8em}}\) How many different arrangements are there of the 11 letters in the word REQUIREMENT?
\(\\[1pt]\)
\({\small(\textrm{b}).\hspace{0.8em}}\) How many different arrangements are there of the 11 letters in the word REQUIREMENT in which the two Rs are together and the three Es are together?
\(\\[1pt]\)
\({\small(\textrm{c}).\hspace{0.8em}}\) How many different arrangements are there of the 11 letters in the word REQUIREMENT in which there are exactly three letters between the two Rs?
\(\\[1pt]\)
Five of the 11 letters in the word REQUIREMENT are selected.
\(\\[1pt]\)
\({\small(\textrm{d}).\hspace{0.8em}}\) How many possible selections contain at least two Es and at least one R?
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/ClhmagqNj_9/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 3.\enspace}\)
9709/52/F/M/20 – Paper 52 March 2020 Probability and Statistics I No 1 \(\\[1pt]\)
The 40 members of a club include Ranuf and Saed. All 40 members will travel to a concert. 35 members will travel in a coach and the other 5 will travel in a car. Ranuf will be in the coach and Saed will be in the car.
\(\\[1pt]\)
In how many ways can the members who will travel in the coach be chosen?
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/CPE1Ybggz1I/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 4.\enspace}\)
9709/52/F/M/20 – Paper 52 March 2020 Probability and Statistics I No 4 \(\\[1pt]\)
Richard has 3 blue candles, 2 red candles and 6 green candles. The candles are identical apart from their colours. He arranges the 11 candles in a line.
\(\\[1pt]\)
\({\small(\textrm{a}).\hspace{0.8em}}\) Find the number of different arrangements of the 11 candles if there is a red candle at each end.
\(\\[1pt]\)
\({\small(\textrm{b}).\hspace{0.8em}}\) Find the number of different arrangements of the 11 candles if all the blue candles are together and the red candles are not together.
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/CPRvradHL0m/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too! \(\\[1pt]\)
\({\small 5.\enspace}\)
9709/61/M/J/19 – Paper 61 June 2019 Probability and Statistics I No 8 \(\\[1pt]\)
Freddie has 6 toy cars and 3 toy buses, all different. He chooses 4 toys to take on holiday with him.
\(\\[1pt]\)
\({\small(\textrm{i}).\hspace{0.8em}}\) In how many different ways can Freddie choose 4 toys?
\(\\[1pt]\)
\({\small(\textrm{ii}).\hspace{0.7em}}\) How many of these choices will include both his favourite car and his favourite bus?
\(\\[1pt]\)
Freddie arranges these 9 toys in a line.
\(\\[1pt]\)
\({\small(\textrm{iii}).\hspace{0.6em}}\) Find the number of possible arrangements if the buses are all next to each other.
\(\\[1pt]\)
\({\small(\textrm{iv}).\hspace{0.6em}}\) Find the number of possible arrangements if there is a car at each end of the line and no buses
are next to each other.
Check out my solution here:
\(\\[1pt]\)
https://www.instagram.com/p/CP3OHYDD4Sj/ \(\\[1pt]\)
Follow my instagram for daily updates on many more solutions on other topics too!\(\\[1pt]\)
PRACTICE MORE WITH THESE QUESTIONS BELOW!
\({\small 1.\enspace}\) In a 60-metre hurdles race there are five runners, one from each of the nations Austria, Belgium, Canada, Denmark and England.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) How many different finishing orders are there?
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}\) What is the probability of predicting the finishing order by choosing first, second, third, fourth and fifth at random?
\({\small 2. \enspace}\) In a ‘Goal of the season’ competition, participants are asked to rank ten goals in order of quality.
The organisers select their ‘correct’ order at random. Anybody who matches their order will be invited to join the television commentary team for the next international match.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) What is the probability of a participant’s order being the same as that of the organisers?
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}\) Five million people enter the competition. How many people would be expected to join the commentary team?
\({\small 3. \enspace}\) How many arrangements of the word ACHIEVE are there if
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.8em}}\) there are no restrictions on the order the letters are to be in
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.7em}}\) the first letter is an A
\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.5em}}\) the letters A and I are to be together.
\({\small\hspace{1.2em} (\textrm{iv}).\hspace{0.6em}}\)the letters C and H are to be apart.
\({\small 4. \enspace (\textrm{i}).\hspace{0.7em}}\) A football team consists of 3 players who play in a defence position, 3 players who play in a midfield position and 5 players who play in a forward position. Three players are chosen to collect a gold medal for the team. Find in how many ways this can be done
\({\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}\) if the captain, who is a midfield player, must be included, together with one defence and one forward player.
\({\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}\) if exactly one forward player must be included, together with any two others.
\({\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find how many different arrangements there are of the nine letters in the words GOLD MEDAL
\({\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}\) if there are no restrictions on the order of the letters,
\({\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}\) if the two letters D come first and the letters L come last.
\({\small 5. \enspace}\) The diagram shows the seating plan for passengers in a minibus, which has 17 seats arranged in 4 rows. The back row has 5 seats and the other 3 rows have 2 seats on each side. 11 passengers get on the minibus.
\(\\[1pt]\)
\(\\[1pt]\)
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.7em}}\) How many possible seating arrangements are there for the 11 passengers?
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) How many possible seating arrangements are there if 5 particular people sit in the back row?
Of the 11 passengers, 5 are unmarried and the other 6 consist of 3 married couples.
\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.5em}}\) In how many ways can 5 of the 11 passengers on the bus be chosen if there must be 2 married couples and 1 other person, who may or may not be married?
\({\small 6. \enspace}\) A choir consists of 13 sopranos, 12 altos, 6 tenors and 7 basses. A group consisting of 10 sopranos, 9 altos, 4 tenors and 4 basses is to be chosen from the choir.
\({\small\hspace{1.2em} (\textrm{i}).\hspace{0.7em}}\) In how many different ways can the group be chosen?
\({\small\hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) In how many ways can the 10 chosen sopranos be arranged in a line if the 6 tallest stand next to each other?
\({\small\hspace{1.2em} (\textrm{iii}).\hspace{0.5em}}\) The 4 tenors and the 4 basses in the group stand in a single line with all the tenors next to each other and all the basses next to each other. How many possible arrangements are there if three of the tenors refuse to stand next to any of the basses?
\({\small 7. \enspace (\textrm{i}).\hspace{0.7em}}\) Find how many numbers between 5000 and 6000 can be formed from the digits 1, 2, 3, 4, 5 and 6
\({\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}\) if no digits are repeated,
\({\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}\) if repeated digits are allowed.
\({\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the number of ways of choosing a school team of 5 pupils from 6 boys and 8 girls
\({\small\hspace{2.8em}(\textrm{a}).\hspace{0.7em}}\) if there are more girls than boys in the team,
\({\small\hspace{2.8em}(\textrm{b}).\hspace{0.7em}}\) if three of the boys are cousins and are either all in the team or all not in the team.
\({\small 8. \enspace (\textrm{i}).\hspace{0.7em}}\) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that all four Es are together.
\({\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the number of different ways in which all 12 letters of the word STEEPLECHASE can be arranged so that the Ss are not next to each other.
Four letters are selected from the 12 letters of the word STEEPLECHASE.
\({\small \hspace{1.2em} (\textrm{iii}).\hspace{0.4em}}\) Find the number of different selections if the four letters include exactly one S.
\({\small 9. \enspace (\textrm{i}).\hspace{0.7em}}\) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that all three Os are together and both Ts are together.
\({\small \hspace{1.2em} (\textrm{ii}).\hspace{0.6em}}\) Find the number of different ways in which the 9 letters of the word TOADSTOOL can be arranged so that the Ts are not together.
\({\small \hspace{1.2em} (\textrm{iii}).\hspace{0.4em}}\) Find the probability that a randomly chosen arrangement of the 9 letters of the word TOADSTOOL has a T at the beginning and a T at the end.
\({\small \hspace{1.2em} (\textrm{iv}).\hspace{0.6em}}\) Five letters are selected from the 9 letters of the word TOADSTOOL. Find the number of different selections if the five letters include at least 2 Os and at least 1 T.
\({\small 10. \enspace (\textrm{a}).\hspace{0.7em}}\) A group of 6 teenagers go boating. There are three boats available. One boat has room for 3 people, one has room for 2 people and one has room for 1 person. Find the number of different ways the group of 6 teenagers can be divided between the three boats.
\({\small \hspace{1.6em} (\textrm{b}).\hspace{0.6em}}\) Find the number of different 7-digit numbers which can be formed from the seven digits 2, 2, 3, 7, 7, 7, 8 in each of the following cases.
\({\small \hspace{2.8em} (\textrm{i}).\hspace{0.7em}}\) The odd digits are together and the even digits are together.
\({\small \hspace{2.8em} (\textrm{ii}).\hspace{0.6em}}\) The 2s are not together.
As always, if you have any particular questions to discuss, leave it in the comment section below. Cheers =) .
A pharmaceutical company is researching into a new malaria drug. The team to carry out the research is to consist of 6 member staff. If there are 8 biochemist and 10 microbiologist in the company, in how many ways can this be done?
(A) if the team can be constituted anyhow.
(B) if the team must include at least a biochemist and microbiologist.
(C) if the only chief technologist is a biochemist and must be in the team.
I have a question about the answer to 5a problem 6; the envelope problem. If the total number of possible ways to fill the five addressed envelopes is 5! = 120, and the question asked “how many ways can the letters be placed in each envelope without getting every letter in the right envelope” then shouldn’t that one possible way of getting them placed right be subtracted from that total, leaving the answer 120 – 1 = 119 ? Please help me understand the why the answer remains 120? Thank you!
Can u solve Exercises 6, 7 and 8 from this book Advanced level Mathematics Statistics 1
how many 0dd number greater than 300 can be formed using the digit 1,2,3,4,5.
a)without repetition
b)with repetition
You can solve this question fairly easily. Firstly, count the possible 3-digit odd numbers.
Next, count the possible 4-digit odd numbers and then count the possible 5-digit odd numbers.
Cheers
Mr. Will
Could you please upload the answers of “Practice Questions” so that I can check my answers.
Can u solve Exercises 6, 7 and 8 from this book Advanced level Mathematics Statistics 1
yes
please share me guide book
Asslamo Aliakum sir I hopes you are well
i want 5 to on word chapter like this can you send me any link
when i click to show answer they blank page is opened
Just part of an adsense, close the page and see the solution.
Cheers,