Permutations & Combinations Questions and Answers

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.

Answer: C

Count all groups of three, then remove the forbidden ones.

\[ \text{total} = {}^{5}C_{3} = 10 \]

Groups containing both C and D need one more from A, B, E:

\[ {}^{3}C_{1} = 3 \]

\[ 10 - 3 = 7 \]

Counting the forbidden case and subtracting is quicker than enumerating the seven allowed groups directly.

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.

Answer: D

A line needs two points. Two points on the same line give nothing new - they just re-describe that line.

One point from each line:

\[ 6 \times 5 = 30 \text{ new lines} \]

The two original lines add 2 more.

\[ 30 + 2 = 32 \]

Taking \( {}^{11}C_{2} = 55 \) would count every pair, including the \( {}^{6}C_{2} + {}^{5}C_{2} = 25 \) collinear pairs that produce no new line - and \( 55 - 25 + 2 = 32 \) again.

Answer: B

Each digit runs from 0 to 8, giving nine choices, and repetition is allowed.

\[ 9 \times 9 \times 9 = 729 \text{ combinations} \]

\[ 729 \times 3 = 2187 \text{ seconds} \]

\[ \frac{2187}{60} = 36.45 \text{ minutes} \]

The digits go up to 8, not 9, so it is \( 9^{3} \) and not \( 10^{3} \).