Use the identity $(1+x)^{2n} = (1+x)^{n} (1+x)^{n}$ to show that $$inom{2n}{0} + inom{2n}{1} + \cdots + inom{2n}{n}^2,$$ where $n$ is a positive integer - HSC - SSCE Mathematics Extension 1 - Question 14 - 2020 - Paper 1

To form a group with an equal number of men and women, we can let the number of men be $k$ (and thus, the number of women will also be $k$). The total number of members in the group will then be $2k$.

The number of ways to select $k$ men from $n$ is given by inom{n}{k}, and similarly for women.

Therefore, the total number of ways to select $k$ men and $k$ women is: $\binom{n}{k}\cdot\binom{n}{k} = \left(\binom{n}{k}\right)^2.$

Now, since $k$ can take values from 0 to $n$, we sum over all possible $k$: $\sum_{k=0}^{n} \left(\binom{n}{k}\right)^2,$ which, using a combinatorial identity, equals $\binom{2n}{n}$. Thus, the total number of ways to do this is $\binom{2n}{n}$.

To select leaders from the group chosen, we first need to have $2m$ members where $m$ is the number of pairs $(1 ext{ man}, 1 ext{ woman})$ from part (ii).

For each of the $k$ pairs, we can choose one man in $k$ ways and one woman in $k$ ways, which leads to: $k \cdot k = k^2.$

Thus, the total ways to choose leaders is: $k^2 = \binom{n}{1}^2 + \binom{n}{2}^2 + ... + \binom{n}{n}^2.$

Therefore, the total number of ways to choose the even number of people and then the leaders is: $i^2 \binom{n}{1} + 2^2 \binom{n}{2} + \cdots + n^2 \binom{n}{n}^2.$

When reversing the process, we first choose the leaders (1 man and 1 woman), then we proceed to select an equal number of remaining members.

Choosing one man can be done in $n$ ways, and choosing one woman can also be done in $n$ ways. Thus, the initial choices of leaders account for: $n^2.$

Then we apply part (ii) to the remaining $2(n-1)$ members: $\binom{2(n-1)}{(n-1)}.$

Thus the total number of ways to select the even number of people is: $n^2 \cdot \binom{2(n-1)}{n-1}.$