Permutations & Combinations Questions and Answers

Q: A group consists of 4 men, 6 women and 5 children. In how many ways can 3 men, 2 women and 3 children selected from the given group?

\[ {}^{4}C_{3} \times {}^{6}C_{2} \times {}^{5}C_{3} \]\[ = 4 \times 15 \times 10 = 600 \]Each group is chosen from its own pool, so the three counts are independent and multiply.

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81.

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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82.

In a group containing 6 cows and 4 buffalos, 4 livestock are to be selected in such a way that at least 1 cow should always be present. How many way of doing that are possible?

A.209
B.205
C.194
D.163

Answer: A

There are ten animals in all. "At least one cow" is easiest through the complement.

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

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

\[ 210 - 1 = 209 \]

With only four buffalos there is exactly one all-buffalo selection, which is what makes the complement so small.

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83.

A group consists of 4 men, 6 women and 5 children. In how many ways can 3 men, 2 women and 3 children selected from the given group?

A.300
B.450
C.600
D.750

Answer: C

\[ {}^{4}C_{3} \times {}^{6}C_{2} \times {}^{5}C_{3} \]

\[ = 4 \times 15 \times 10 = 600 \]

Each group is chosen from its own pool, so the three counts are independent and multiply.

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84.

There are 35 people in a group. There are 12 school girls, 10 school boys, 5 senior citizens and 8 babies in the group. The organizer of the group wants to select a school girl or a school boy as leader of the group. In how many ways can he do so?

A.22
B.1/2
C.13
D.35

Answer: A

The leader is to be a school girl or a school boy, and no one is both, so the two counts simply add.

\[ 12 + 10 = 22 \]

"Or" between mutually exclusive choices means addition; it would be multiplication only if a girl and a boy were both being chosen.

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85.

Three flags each of different colours are available for a military exercise, Using these flags different codes can be generated by waving

I. Single flag of different colours

II. Any two flags in a different sequence of colours.

III. three flags in a different sequence of colours.

The maximum number of codes that can be generated is.

A.6
B.9
C.15
D.18

Answer: C

This type of question becomes very easy when we assume three colour are red(R) blue(B) and Green(G).

Now according to the statement 1 i.e.., codes can be generated by waving single flag of different colours, then number of ways are three i.e.., R.B.G from statement III three flags in different sequence of colours, then number of ways are six i.e.., RBG, BGR, GBR, RGB, BRG, GRB.

Hence total number of ways by changing flag = 3+ 6 +6 = 15

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86.

In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?

A.10080
B.4989600
C.120960
D.None of these

Answer: C

MATHEMATICS has eleven letters, with M, A and T each appearing twice. Its vowels are A, E, A, I and its consonants M, T, H, M, T, C, S.

Treating the four vowels as one block gives eight units, among which M and T each repeat:

\[ \frac{8!}{2!\,2!} = \frac{40320}{4} = 10{,}080 \]

Within the block the four vowels include two A's:

\[ \frac{4!}{2!} = 12 \]

\[ 10{,}080 \times 12 = 120{,}960 \]

Three separate repetitions have to be divided out - the M's and T's among the units, and the A's inside the block.

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87.

A two number committee comprising of one male and the female number is to be constituted out of five male and 3 females. Amongst the females, Ms. A refused to be number committee in which Mr.B is taken as the number of how many different ways can the committee be constituted?

A.11
B.30
C.14
D.20

Answer: C

For each visualization let us name the females and Females(3) Males(5)

A B C D E F G H

Since A cannot go with B.

She will make team with four males in four ways AD,AF,AG,AH.

Since there is no compulsion with females C and E.

They can make team with 5 males in 5 different ways each.

Therefore, Total number of ways = 4+5+5 =14

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88.

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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