There are $5$ balls of different colors and $5$ boxes of colors same as those of the balls.In how many ways in which the balls one in each box could be placed such that a ball does not go to the box of same color?

44

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53

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81

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95

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Solution

The correct option is A $44$
Darrangement of $5$ balls and $5$ boxes
No ball to be in its respective box
$= [1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}] 5! = [\frac{60 - 20 + 5 - 1}{120}] \times 120 = 44$
Therefore number of ways $= 44$

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There are 5 balls of different colors and 5 boxes of colors same as those of the balls.In how many ways in which the balls one in each box could be placed such that a ball does not go to the box of same color?

The correct option is A 44Darrangement of 5 balls and 5 boxes No ball to be in its respective box =[1−11!+12!−13!+14!−15!]5!=[60−20+5−1120]×120=44Therefore number of ways =44

There are $5$ balls of different colors and $5$ boxes of colors same as those of the balls.In how many ways in which the balls one in each box could be placed such that a ball does not go to the box of same color?

The correct option is A $44$
Darrangement of $5$ balls and $5$ boxes
No ball to be in its respective box
$= [1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!}] 5! = [\frac{60 - 20 + 5 - 1}{120}] \times 120 = 44$
Therefore number of ways $= 44$