Q51
ArithmeticPermutations, Combinations & Probability
A clerk is given 3 letters written in a language unknown to her. She is also given an envelope with address corresponding to each letter written in same language. If the clerk has to put the letters in the envelopes, in how many ways can she do it so that none of the 3 envelopes has the letter with the correct address in it?
- a.1
- b.2
- c.3
- d.6
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.