Permutations, Combinations & Probability: previous-year questions

Answer: (B) 2

This is a derangement problem: arrangements in which no object occupies its own place. For three letters and three envelopes there are $3! = 6$ arrangements in all, and only two of them, (2, 3, 1) and (3, 1, 2), put every letter in a wrong envelope. The general formula is $D_n = n!\left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots \pm \frac{1}{n!}\right)$, which gives $D_2 = 1$, $D_3 = 2$, $D_4 = 9$ and $D_5 = 44$, values worth memorising. Option D (6) counts all arrangements, and option A (1) is the number in which every letter is correctly placed.

Answer: (C) 13

This is the pigeonhole principle: with 12 months (pigeonholes), 12 members could all have different birth months, but a 13th member must share a month with someone. So the minimum that guarantees a match is $12 + 1 = 13$. Option B (12) is the largest group that can still avoid a match, and option D (24) is a distractor based on 'two per month'. Extension often asked: to guarantee that at least three members share a birth month you need $2 \times 12 + 1 = 25$; to guarantee two people with the same birthday you need 367 people.

Answer: (A) 10

${}^nP_4=n(n-1)(n-2)(n-3)=5040$, a product of four consecutive integers. Trying $n=10$: $10\times9\times8\times7=5040$, so $n=10$. Check against the other options: for $n=12$ the product is $12\times11\times10\times9=11880$, already too large, and $n=16$ or $20$ would be larger still. A handy fact is that $5040=7!$, and $7!=10\times9\times8\times7$ because $6\times5\times4\times3\times2=720=10\times9\times8$.

Answer: (B) 15

Every match needs a pair of players, so the number of matches is the number of ways of choosing 2 players out of 6: $\binom{6}{2} = \frac{6 \times 5}{2} = 15$. General rule for a single round-robin with $n$ players: $\frac{n(n-1)}{2}$ matches. Option C (30) would be right only if every pair played twice (double round-robin), i.e. $n(n-1)$.

Answer: (B) 110

Captain can be chosen in 11 ways; vice-captain from remaining 10 in 10 ways. Total = $11 \times 10 = 110$.