Permutations & Combinations Questions and Answers

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.

Answer: B

Four-digit numbers from five distinct digits, no repetition:

\[ {}^{5}P_{4} = 120 \text{ numbers} \]

By symmetry each digit occupies each position equally often:

\[ \frac{120}{5} = 24 \text{ times per position} \]

The digits sum to \( 1+3+5+7+9 = 25 \), so each position contributes \( 24 \times 25 = 600 \) to its place value.

\[ \text{sum} = 600 \times (1000 + 100 + 10 + 1) = 600 \times 1111 \]

\[ = 666{,}600 \]

The symmetry argument avoids listing 120 numbers: no digit is favoured in any column.

Answer: C

GEOMETRY has eight letters with E appearing twice. Its vowels are E, O, E.

Treat the three vowels as one block, giving six units to arrange (the block plus G, M, T, R, Y):

\[ 6! = 720 \]

Within the block the three vowels can be ordered in

\[ \frac{3!}{2!} = 3 \text{ ways} \]

since the two E's are identical.

\[ 720 \times 3 = 2160 \]

Treating the E's as distinguishable would give \( 720 \times 6 = 4320 \) - double-counting every arrangement, because swapping the two E's changes nothing visible.

Answer: D

Count all arrangements, then remove those with a same-coloured pair adjacent.

\[ \text{total} = \frac{6!}{2!\,2!\,2!} = \frac{720}{8} = 90 \]

Treating one colour's pair as a single block:

\[ \frac{5!}{2!\,2!} = 30 \quad \text{for each of the 3 colours} \]

Two colours blocked:

\[ \frac{4!}{2!} = 12 \quad \text{for each of the 3 pairs of colours} \]

All three blocked: \( 3! = 6 \).

By inclusion–exclusion the number with at least one adjacent pair is

\[ 3(30) - 3(12) + 6 = 90 - 36 + 6 = 60 \]

\[ 90 - 60 = 30 \]

All three correction terms are needed; stopping after the first subtraction would leave zero.

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.