Permutations & Combinations Questions and Answers

Answer: A

ENCYCLOPAEDIA has thirteen letters: vowels E, O, A, E, A, I (six, with E and A each twice) and consonants N, C, Y, C, L, P, D (seven, with C twice).

Positions 2, 4, 6, 8, 10 and 12 are the six even ones - exactly enough for the vowels.

\[ \text{vowels} = \frac{6!}{2!\,2!} = \frac{720}{4} = 180 \]

\[ \text{consonants in the 7 odd positions} = \frac{7!}{2!} = 2520 \]

\[ 180 \times 2520 = 453{,}600 \]

The counts of vowels and even positions match exactly, which is what makes the arrangement possible at all.

Answer: B

INDIA has five letters with I twice, and its vowels are I, I, A.

\[ \text{total} = \frac{5!}{2!} = 60 \]

With the three vowels as one block there are three units (block, N, D):

\[ 3! \times \frac{3!}{2!} = 6 \times 3 = 18 \text{ with the vowels together} \]

\[ 60 - 18 = 42 \]

The repeated I forces a division by \( 2! \) in both counts.

Answer: D

The two choices are independent, so they multiply.

\[ 9 \times 12 = 108 \]

Each of the nine trousers can be paired with each of the twelve shirts. Adding them would answer a different question - how many garments he could pick one of.

Answer: D

TIME has four distinct letters, and order matters in a word.

\[ {}^{4}P_{3} = 4 \times 3 \times 2 = 24 \]

The corresponding combination \( {}^{4}C_{3} = 4 \) would count only which three letters are used, not how they are arranged.

Answer: B

The first and last letters are fixed as N and A, leaving O, K and I for the three middle positions.

\[ 3! = 6 \]

Without the restriction all five letters could move, giving \( 5! = 120 \).

Answer: B

MEADOWS has seven distinct letters: vowels E, A, O and consonants M, D, W, S.

The even places among positions 1 to 7 are 2, 4 and 6 - three places for three vowels.

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

\[ \text{consonants in the 4 odd places} = 4! = 24 \]

\[ 6 \times 24 = 144 \]

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

Answer: B

The two selections are independent, so their counts multiply.

\[ {}^{5}C_{2} \times {}^{6}C_{3} = 10 \times 20 = 200 \]

Order does not matter within either group, so combinations are used rather than permutations.

Answer: C

One particular man and one particular woman are fixed in the group, so only the rest are free to choose.

\[ \text{2 more men from the remaining 4} = {}^{4}C_{2} = 6 \]

\[ \text{1 more woman from the remaining 5} = {}^{5}C_{1} = 5 \]

\[ 6 \times 5 = 30 \]

Fixing members always reduces both the pool and the number still to be chosen.

Answer: A

There are \( 5 + 6 = 11 \) people, and eight are to be chosen with no restriction on the mix.

\[ {}^{11}C_{8} = {}^{11}C_{3} = \frac{11 \times 10 \times 9}{6} = 165 \]

Choosing 8 to include is the same as choosing 3 to leave out, which makes the arithmetic much lighter.

Answer: D

Eighteen stations lie between the two named ones, so there are 20 stations in all.

A ticket names both a source and a destination, so order matters:

\[ {}^{20}P_{2} = 20 \times 19 = 380 \]

The 190 on offer counts station pairs rather than tickets, missing the return direction.