Supposing $4$ letters are placed in $4$ different envelopes. In how many ways can they be taken out from their original envelopes and distributed among the $4$ different envelopes so that no letter remains in its original envelope?

7

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8

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9

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None of these

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Solution

The correct option is D $9$
Using the formula for the number of derangement that are possible out of $4$ letters in $4$ envelopes, we get the number of ways as
$4! (1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!})$
= $24 (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24})$
= $24 (1 - 1 + \frac{1 (12)}{2 (12)} - \frac{1 (4)}{6 (4)} + \frac{1}{24})$
= $24 (\frac{12 - 4 + 1}{24})$
= $24 \times \frac{9}{24}$
= $9$

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Supposing 4 letters are placed in 4 different envelopes. In how many ways can they be taken out from their original envelopes and distributed among the 4 different envelopes so that no letter remains in its original envelope?

The correct option is D 9Using the formula for the number of derangement that are possible out of 4 letters in 4 envelopes, we get the number of ways as 4!(1−1+12!−13!+14!) = 24(1−1+12−16+124) = 24(1−1+1(12)2(12)−1(4)6(4)+124) =24(12−4+124) = 24×924 =9

Supposing $4$ letters are placed in $4$ different envelopes. In how many ways can they be taken out from their original envelopes and distributed among the $4$ different envelopes so that no letter remains in its original envelope?