There are 4 distinct colored balls and 4 boxe | vedclass

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(C) This is a problem of derangements,denoted by $D_n$ or $!n$.For $n = 4$ objects,the number of derangements is given by the formula:$D_n = n! \left(

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) This is a problem of derangements,denoted by $D_n$ or $!n$.For $n = 4$ objects,the number of derangements is given by the formula:$D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \right)$Substituting $n = 4$:$D_4 = 4! \left( 1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$$D_4 = 24 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} \right)$$D_4 = 12 - 4 + 1 = 9$Thus,there are $9$ ways to place the balls such that no ball goes into a box of its own color.

There are $4$ distinct colored balls and $4$ boxes of the same colors as the balls. In how many ways can the balls be placed in the boxes such that no ball goes into a box of its own color?

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