Permutations & Combinations Questions and Answers

Answer: C

The number of ways is 6! / 2.

MIDDLE has 6 letters, with D appearing twice. Arrangements of 6 cards are 6!, but swapping the two identical D cards gives the same visible arrangement, so each is counted twice:

\[ \frac{6!}{2!} = \frac{720}{2} = 360 \]

6! would treat the two Ds as different, 5! fixes one card, and 6! - 4! has no basis.

In general, n objects with one item repeated r times can be arranged in \(\frac{n!}{r!}\) distinct ways.

Answer: A

There are 20 ways.

Choosing one envelope (5 ways) and one stamp (4 ways) are independent choices, so by the multiplication principle the number of ways to buy one of each is

\[ 5 \times 4 = 20 \]

9 is 5 + 4, the count of ways to buy an envelope or a stamp, not both, which is the common slip. 45 and 54 fit no step.

Answer: B

4 people can sit round a circular table in 3! ways.

In a circular arrangement, rotating everyone together gives the same seating, so one person's place is fixed and the rest are arranged relative to them.

\[ (n - 1)! = 3! = 6 \]

4! = 24 counts arrangements in a row, treating rotations as different. 4 and 1 undercount the arrangements.

For n people round a table where clockwise and anticlockwise orders count as the same, such as beads on a necklace, the count would be (n - 1)!/2.

Answer: C

There will be 45 handshakes.

Each handshake involves a pair of people, so count the pairs that can be chosen from 10.

\[ {}^{10}C_{2} = \frac{10 \times 9}{2} = 45 \]

Multiplying 10 by 9 without halving would count every handshake twice (90). 35, 40 and 50 do not correspond to any pairing count of 10 people.

Answer: B

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

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

The 870 on offer is \( 30 \times 29 \), which counts each handshake twice - once from each participant's side.

Answer: A

Every pair of teams meets once, and the order within a pair is irrelevant.

\[ {}^{9}C_{2} = \frac{9 \times 8}{2} = 36 \]

Counting ordered pairs would give 72, treating "A versus B" and "B versus A" as different matches.

Answer: D

SMOKEJACK has nine letters with K appearing twice. J must be one of the four, so three more are chosen from S, M, O, K, E, A, C - with K available twice.

At most one K - choose 3 of the 7 distinct letters:

\[ {}^{7}C_{3} = 35 \]

Both K's - choose 1 more from the other 6:

\[ {}^{6}C_{1} = 6 \]

\[ 35 + 6 = 41 \]

The repeated K is what makes this more than a plain \( {}^{7}C_{3} \); the two-K selections are genuinely different combinations.

Answer: C

There are \( 2 + 3 + 4 = 9 \) chairs. "At least one yellow" is easiest through the complement.

\[ \text{total} = {}^{9}C_{3} = 84 \]

\[ \text{no yellow} = {}^{6}C_{3} = 20 \]

\[ 84 - 20 = 64 \]

Counting exactly one, two and three yellows separately gives \( 3 \times 15 + 3 \times 6 + 1 = 64 \) as well - the same answer for three times the work.

Answer: C

A ticket names a starting stop and a destination, so the two directions need different tickets: the order matters.

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

Treating the journey as unordered would give \( {}^{20}C_{2} = 190 \) - the number of station pairs, not of tickets.

Answer: D

Treat each subject as a single block, then arrange within each block.

\[ \text{blocks among themselves} = 3! = 6 \]

\[ \text{History} = 3! = 6, \quad \text{Science} = 4! = 24, \quad \text{Maths} = 2! = 2 \]

\[ 6 \times 6 \times 24 \times 2 = 1728 \]

Forgetting the \( 3! \) for the blocks themselves would give 288; the subjects can appear in any order on the shelf.