(C) This is a problem of derangements where $n = 4$ items are to be placed in $n$ boxes such that no item goes into its correct box.The number of dera

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

Class 11 Mathematics Permutation and Combination Division into groups, Derangements

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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

IIT 1992 All IIT

A
$8$
B
$7$
C
$9$
D
None of these

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Permutation and Combination

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Division into groups, Derangements

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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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