There are 4 balls of different colors and 4 b | vedclass

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(C) This is a problem of derangements of $4$ objects,denoted as $D_4$.The formula for derangements of $n$ objects is $D_n = n! \left( 1 - \frac{1}{1!}

Vedclass · Accepted Answer

$9$

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(C) This is a problem of derangements of $4$ objects,denoted as $D_4$.The formula for derangements of $n$ objects is $D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!} \right)$.For $n = 4$:$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 = 24 \left( \frac{12 - 4 + 1}{24} \right)$$D_4 = 9$.Thus,there are $9$ ways to place the balls such that no ball is in its own color box.

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

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