Question

Let $$A$$ be a set containing $$n$$ elements. A subset $$P$$ of the set $$A$$ is chosen at random.The set $$A$$ is reconstructed by replacing the element of $$P$$, and another subsets $$Q$$ of $$A$$ is chosen at random. The probability that $$\left( P\cap Q \right)$$ contains exactly $$m(m<n)$$ elements is

A
3nm4n
B
nCm3m4n
C
nCm3nm4n
D
none of these

Solution

The correct option is C $$\displaystyle \frac { _{ }^{ n }{ { C }_{ m } }{ 3 }^{ n-m } }{ { 4 }^{ n } }$$We know that the number of subsets of a set containing $$n$$ elements is $${ 2 }^{ n }$$.Therefore, the number of ways of choosing $$P$$ and $$Q$$ is $$^{ { 2 }^{ n } }{ { C }_{ 1 } }\times _{ }^{ { 2 }^{ n } }{ { C }_{ 1 } }={ 2 }^{ n }\times { 2 }^{ n }={ 4 }^{ n }$$.Out of $$n$$ elements, $$m$$ elements can be chosen in $$^{ n }{ { C }_{ m } }$$ ways.If $$\left( P\cap Q \right)$$ contains exactly in elements, then from the remaining $$n-m$$ elements either an element belongs to $$P$$ or $$Q$$ but not both $$P$$ and $$Q$$.Suppose $$P$$ contains $$r$$ elements from the remaining $$n-m$$ elements.Then $$Q$$ may contain any number of elements from the remaining $$(n-m)-r$$ elements. Therefore, $$P$$ and $$Q$$ can be chosen in $$^{ n-m }{ { C }_{ r } }{ 2 }^{ \left( n-m \right) -r }$$But $$r$$ can vary from $$0$$ to $$(n-m)$$. So, $$P$$ and $$Q$$ can be chosen in general in$$\displaystyle \left( \sum _{ r=0 }^{ n-m }{ ^{ n-m }{ { C }_{ r } }{ 2 }^{ \left( n-m \right) -r } } \right) _{ }^{ n }{ { C }_{ m } }={ \left( 1+2 \right) }^{ n-m }\times _{ }^{ n }{ { C }_{ m } }=_{ }^{ n }{ { C }_{ m } }\times { 3 }^{ n-m }$$Hence, required probability $$\displaystyle =\dfrac { ^{ n }{ { C }_{ m } }\times { 3 }^{ n-m } }{ { 4 }^{ n } }$$Maths

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