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 each in a box, could be placed such that a ball does not go to a box of its own colour is

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Solution

The correct option is B 9
A ball of the particular colour can be distributed in three boxes having colours different from that of the ball
The second ball having a colour same as that of the box in which the first ball is allotted can be distributed in three different boxes because each of the remaining three boxes will have the colours different from the second ball being allotted
Each of the remaining two balls can be distributed only in one way as for each,only one box having different colour will be available.Hence the number of ways of distributing the balls will be equal to $3 \times 3 \times 1 \times 1 = 9$
Hence option B is correct

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