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(C) This is a classic problem of derangements,where $n$ items are placed in $n$ containers such that no item is in its correct container.The number of

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(C) This is a classic problem of derangements,where $n$ items are placed in $n$ containers such that no item is in its correct container.The number of derangements of $n$ objects,denoted by $D_n$,is given by the formula:$D_n = n! \left( 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \dots + (-1)^n \frac{1}{n!} \right)$For $n = 4$:$D_4 = 4! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} \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 = 12 - 4 + 1 = 9$Thus,there are $9$ ways to place the balls such that no ball goes into the box of its own colour.

There are four balls of different colours and four boxes of colours same as those of the balls. The number of ways in which the balls,one in each box,could be placed such that a ball does not go to the box of its own colour is

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