Permutations & Combinations Questions and Answers

Q: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

Choose the letters first, then arrange them.\[ {}^{7}C_{3} \times {}^{4}C_{2} = 35 \times 6 = 210 \text{ selections} \]Each selection of five letters can be arranged in\[ 5! = 120 \text{ ways} \]\[ 210 \times 120 = 25{,}200 \]A "word" is an ordered string, so the arrangement step is essential; stopping at 210 counts only which letters are used.

Q: In how many ways can the letters of the word LEADER be arranged ?

Correct answer: 360The word has 6 letters, of which E occurs twice and the rest once each.\[\frac{6!}{2!} = \frac{720}{2} = 360\]Dividing by 2! removes the arrangements that merely swap the two identical E’s.

Q: In how many ways can the letters of the word 'LEADER' be arranged?

LEADER has six letters with E appearing twice.\[ \frac{6!}{2!} = \frac{720}{2} = 360 \]The division by \( 2! \) removes the double-counting from swapping the two identical E's, which changes nothing visible.

Q: In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

The men and women are drawn from separate pools.\[ {}^{7}C_{5} \times {}^{3}C_{2} = 21 \times 3 = 63 \]\( {}^{7}C_{5} = {}^{7}C_{2} = 21 \), which is the quicker way to evaluate it.

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101.

Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

A.210
B.1050
C.25200
D.21400

Answer: C

Choose the letters first, then arrange them.

\[ {}^{7}C_{3} \times {}^{4}C_{2} = 35 \times 6 = 210 \text{ selections} \]

Each selection of five letters can be arranged in

\[ 5! = 120 \text{ ways} \]

\[ 210 \times 120 = 25{,}200 \]

A "word" is an ordered string, so the arrangement step is essential; stopping at 210 counts only which letters are used.

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102.

In how many ways can the letters of the word LEADER be arranged ?

A.72
B.144
C.360
D.720

Answer: C

Correct answer: 360

The word has 6 letters, of which E occurs twice and the rest once each.

\[\frac{6!}{2!} = \frac{720}{2} = 360\]

Dividing by 2! removes the arrangements that merely swap the two identical E’s.

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103.

In how many ways can the letters of the word 'LEADER' be arranged?

A.72
B.144
C.360
D.720

Answer: C

LEADER has six letters with E appearing twice.

\[ \frac{6!}{2!} = \frac{720}{2} = 360 \]

The division by \( 2! \) removes the double-counting from swapping the two identical E's, which changes nothing visible.

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104.

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?

A.159
B.194
C.205
D.209

Answer: D

There are ten children. "At least one boy" is easiest through the complement.

\[ \text{total} = {}^{10}C_{4} = 210 \]

\[ \text{no boy} = {}^{4}C_{4} = 1 \]

\[ 210 - 1 = 209 \]

With only four girls there is exactly one all-girl selection, so the complement is a single case.

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105.

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?

A.5
B.10
C.15
D.20

Answer: D

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.

The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.

The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.

Required number of numbers = (1 x 5 x 4) = 20.

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106.

In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?

A.266
B.5040
C.11760
D.86400

Answer: C

The men and the women are chosen from separate pools, so the counts multiply.

\[ {}^{8}C_{5} \times {}^{10}C_{6} \]

\[ = 56 \times 210 = 11{,}760 \]

Using \( {}^{8}C_{3} \) and \( {}^{10}C_{4} \) gives the same values, since choosing whom to include is equivalent to choosing whom to leave out.

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107.

A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?

A.32
B.48
C.64
D.96

Answer: C

The box holds \( 2 + 3 + 4 = 9 \) balls, six of them not black.

\[ \text{total} = {}^{9}C_{3} = 84 \]

\[ \text{no black} = {}^{6}C_{3} = 20 \]

\[ 84 - 20 = 64 \]

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108.

In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?

A.32
B.48
C.36
D.60

Answer: C

DETAIL has six distinct letters: vowels E, A, I and consonants D, T, L.

The odd positions are 1, 3 and 5 - three places for three vowels.

\[ \text{vowels} = 3! = 6 \]

\[ \text{consonants in the even positions} = 3! = 6 \]

\[ 6 \times 6 = 36 \]

The two counts match exactly, which is what makes the restriction satisfiable at all.

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109.

In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?

A.63
B.90
C.126
D.45

Answer: A

The men and women are drawn from separate pools.

\[ {}^{7}C_{5} \times {}^{3}C_{2} = 21 \times 3 = 63 \]

\( {}^{7}C_{5} = {}^{7}C_{2} = 21 \), which is the quicker way to evaluate it.

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110.

How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?

A.40
B.400
C.5040
D.2520

Answer: C

LOGARITHMS has ten letters, all different, and a word is an ordered arrangement.

\[ {}^{10}P_{4} = 10 \times 9 \times 8 \times 7 = 5040 \]

The corresponding combination \( {}^{10}C_{4} = 210 \) would count only which four letters are used, ignoring their order.

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