If 5 letters are taken out of 5 different envelopes. In how many ways, can they be reinserted in the envelopes so that no letter goes to its original envelopes?

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Solution

The correct option is B
44

Number of possible de arrangement that are possible out of n letters in n envelopes.

The number of ways= $D_{n} = n! \sum_{n}^{k = 0} \frac{(- 1)^{k}}{k!}$

Number of possible de arrangements that are possible out of 5 letters in 5 envelopes
= $5! (1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!})$
=44

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If 5 letters are taken out of 5 different envelopes. In how many ways, can they be reinserted in the envelopes so that no letter goes to its original envelopes?

The correct option is B 44 Number of possible de arrangement that are possible out of n letters in n envelopes. The number of ways= Dn=n!∑k=0n(−1)kk! Number of possible de arrangements that are possible out of 5 letters in 5 envelopes =5!(1−11!+12!−13!+14!−15!) =44