Question

# 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

A
7
B
8
C
9
D
20

Solution

## The correct option is C $$9$$We know that, if $$n$$ different things are arranged in a row,  then the number of ways in which they can rearranged, so that none of them occupies its original place is $$\displaystyle n!\left( 1-\frac { 1 }{ 1! } +\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +...+\frac { { \left( -1 \right) }^{ n } }{ n! } \right) .$$Now, assume that each ball is placed in the box of its own color and apply the above result. Hence, the required number of different ways is. $$\displaystyle 4!\left( 1-\frac { 1 }{ 1! } +\frac { 1 }{ 2! } -\frac { 1 }{ 3! } +\frac { 1 }{ 4! } \right) =\frac { 4! }{ 2! } -\frac { 4! }{ 3! } +1$$$$=12-4+1=9$$ Mathematics

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