Question

$A$ is a set containing $n$ different elements. A subset $P$ of $A$ is chosen. The set $A$ is reconstructed by replacing the elements of $P$. A subset $Q$ of $A$ is again chosen. The number of ways of choosing $P$ and $Q$ so that $P\cap Q$ contains exactly two elements is:

• A. ${}^n{C}_3\times 2^n$
• B. ${}^n{C}_2\times 3^{n-2}$
• C. $3^{n-2}$
• D. None of these

Answer 1

The correct option is B ${}^n{C}_2\times 3^{n-2}$

Set $A$ has $n$ elements

Also, subset $P$ and subset $Q$ have exactly $2$ common elements.

It means if we remove those two elements, then subset $P$ and $Q$ will have no common elements.

Ways of selecting $2$ elements from $n$ elements $={}^n{C}_2$

Let subset $P$ contains $a$ elements where $0\le a\le n-2$.

Then subset $B$ must be chosen from $(n-2-a)$ elements.

Suppose subset $P$ contains $1$$element,thenpossiblewaysofselectingsubset$P$willbe${}^{ n-1 }{ C }_{1}$

And possible ways of selecting subset $Q$ from remaining $n-3$ elements $=2^{n-3}$

So total possibility if subset $P$ contains $1$ element $={}^{n-2}{C}_1\times 2^{n-3}$

Similarly for $a=0,a=2,a=\mathrm{3...............}a=n-2$

Combining all situation ${}^n{C}_2\times ({}^{n-2}{C}_0\times 2^{n-2}+{}^{n-2}{C}_1\times 2^{n-3}...............{}^{n-2}{C}_{n-2}\times 2^0)={{}^n{C}_2\times (1+2)}^{n-2}={}^n{C}_2\times 3^{n-2}$