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

Question

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

more over the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can

be done is

TestHub · Accepted Answer

Total number of placements of 6 cards into 6 envelopes when no card is placed in the envelope bearing the same number.
$= 6! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!})$
$= 6! (\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} + \frac{1}{6!})$
$= 360 - 120 + 30 - 6 + 5 = 265$
Number of ways the card numbered 1 is placed among the above placements = 5 (envelope number 2, 3, 4, 5, 6)
So the required number of ways when card 1 is placed in the envelope numbered 2 = 265 / 5 = 53

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