There are $n$ men and $n$ women
Now we can select $1$ woman from $n$ in $^{n}C_{1}$
Same way $1$ man can be selected $^{n}C_{1}$ ways
So, for $1$ woman and $1$ man we can get $^{n}C_{1}\times ^{n}C_{1}$ ways $\qquad \to (1)$
Similarly, we can select $2$ woman from $n$ women in $^{n}C_{2}$
$2$ man can be selected in $^{n}C_{2}$ ways
So, for $2$ woman and $2$ man we can get $^{n}C_{2}\times ^{n}C_{2}$ ways $\qquad \to (2)$
$\vdots$
For $n$ woman and $n$ man we can get $^{n}C_{n}\times ^{n}C_{n}$ ways$\qquad \to (n)$
Now, by adding these equations $(1),(2), \dots, (n)$ we get ,
$^{n}C_{0}\times ^{n}C_{0} + ^{n}C_{1}\times ^{n}C_{1}+ ^{n}C_{2}\times ^{n}C_{2} + ^{n}C_{3}\times ^{n}C_{3}+\ldots +^{n}C_{n}\times ^{n}C_{n} =\left(^{2n}C_{n}\right)$
Hence, Ans will be (D).