First slide

Combinations

Question

There are 4 balls of different colours and 4 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 each ball does not go to a box of its own colour, is

Easy

A

5
B

6
C

9
D

12

Solution

The desired number is $4! (\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!}) = 12 - 4 + 1 = 9$

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