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…

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

Aicharya · 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

Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that…

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