A probability problem by Rajdeep Bharati
Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number and moreover the card numbered 1 is always placed in envelope numbered 2.
Find the number of ways to do so.
The answer is 53.
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1 solution
There are 2 possibilities: (i) If the card number ‘2’ goes in the envelope ‘1’ then it is derangement of 4 things which can be done in 4!(1/2! - 1/3! + 1/4!) = 9 ways. (ii) If card number ‘2’ doesn’t go in the envelope ‘1’, then it is derangement of 5 things which can be done in 5!(1/2! - 1/3! - 1/4! + 1/5!) = 44 ways. Hence, total 53 ways are there.