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(D) This is a problem of derangements,denoted by $D_n$,where $n$ is the number of items.For $n=4$,the number of derangements is given by the formula $

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$9$

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(D) This is a problem of derangements,denoted by $D_n$,where $n$ is the number of items.For $n=4$,the number of derangements is given by the formula $D_n = n! 	imes \left(1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}ight)$.Substituting $n=4$:$D_4 = 24 	imes \left(1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}ight)$$D_4 = 24 	imes \left(\frac{12 - 4 + 1}{24}ight)$$D_4 = 24 	imes \frac{9}{24} = 9$.Alternatively,using the inclusion-exclusion principle:Total ways $= 4! = 24$.Ways where at least one letter is in the correct envelope $= \binom{4}{1} 	imes 3! - \binom{4}{2} 	imes 2! + \binom{4}{3} 	imes 1! - \binom{4}{4} 	imes 0! = 24 - 12 + 4 - 1 = 15$.Required ways $= 24 - 15 = 9$.

The number of ways in which $4$ letters can be put into $4$ addressed envelopes such that no letter goes into the envelope meant for it is:

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