Permutations & Combinations Questions and Answers

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: C

For each problem the candidate has four options: attempt one of the three internal choices, or leave the problem alone.

\[ 4^{5} = 1024 \]

One of those is attempting nothing at all, which the question excludes:

\[ 1024 - 1 = 1023 \]

The 511 on offer comes from \( 2^{9} - 1 \) and ignores the internal choices.

Answer: A

The two cases are mutually exclusive, so their counts add.

3 teachers and 3 boys:

\[ {}^{5}C_{3} \times {}^{4}C_{3} = 10 \times 4 = 40 \]

2 girls and 4 teachers:

\[ {}^{3}C_{2} \times {}^{5}C_{4} = 3 \times 5 = 15 \]

\[ 40 + 15 = 55 \]

Both cases give exactly six members, as required, and no group satisfies both conditions at once - so nothing is double-counted.

Answer: C

Treat the five boys as a single block, giving four units to arrange - the block and the three girls.

\[ 4! = 24 \]

Within the block the boys can be ordered in

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

\[ 24 \times 120 = 2880 \]

The block itself may sit anywhere among the girls, which is what the \( 4! \) counts.

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

(i) Two particular members must be included. They take two of the four places, so two more come from the remaining six:

\[ {}^{6}C_{2} = 15 \]

(ii) Two particular members must not be included. All four come from the remaining six:

\[ {}^{6}C_{4} = 15 \]

The two answers coincide because \( {}^{6}C_{2} = {}^{6}C_{4} \) - choosing two to include is the same as choosing four to leave out.

Answer: C

First player can post greeting cards to the remaining 54 players in 54 ways. Second player can post greeting card to the 54 players. Similarly, it happens with the rest of the players. The total numbers of greeting cards posted are
54 + 54 + 54 …
54 (55times) = 54 x 55 = 2970.

Answer: D

A handshake involves two people and the order does not matter, so the count is a combination.

\[ {}^{30}C_{2} = \frac{30 \times 29}{2} = 435 \]

Each of the 30 people shakes 29 hands, but that counts every handshake twice, hence the division by 2.

Answer: B

MATERIAL has eight letters with A twice: vowels A, E, I, A and consonants M, T, R, L.

Treat the four vowels as one block, giving five units:

\[ 5! = 120 \]

Within the block the vowels arrange in

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

\[ 120 \times 12 = 1440 \]

The division by \( 2! \) accounts for the two identical A's; without it the count would double to 2880.

Answer: B

The outward journey has 7 options; the return must avoid whichever was used, leaving 6.

\[ 7 \times 6 = 42 \]

Without the restriction it would be \( 7 \times 7 = 49 \) - the first option, and the difference is exactly the 7 forbidden repeats.