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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
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B
8
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C
9
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D
20
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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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