Team of Four

This problem can be split into 2 parts, the boys and the girls.

Boys

There are 22 boys and only 2 spots for them. For this we can utilise $nCr$ where we are picking a subset of 2 (r) from a set of 22 (n) objects and, as we're picking a team, we do not care about the order. $\begin{pmatrix} 22 \\ 2 \end{pmatrix}=\frac { 22! }{ (22-2)!\cdot 2! } =\frac { 22\times 21\times ...\times 2\times 1 }{ 20\times 19\times ...\times 2\times 1\cdot 2\times 1 } =\frac { 22\times 21 }{ 2 } =231$

Girls

Now, we can do a similar strategy for the girls, using 14 and 2 as our value of n and r: $\begin{pmatrix} 14 \\ 2 \end{pmatrix}=\frac { 14! }{ (14-2)!\cdot 2! } =\frac { 14\times 13\times ...\times 2\times 1 }{ 12\times 11\times ...\times 2\times 1\cdot 2\times 1 } =\frac { 14\times 13 }{ 2 } =91$

Finally we need to multiply them together: $\begin{pmatrix} 22 \\ 2 \end{pmatrix}\begin{pmatrix} 14 \\ 2 \end{pmatrix}=231\times 91=\boxed { 21021 }$

Out of 22 boys 2 boys could be selected in 22C2 ways and same applied for girls which is 14C2 ways.... So at a total, it could be selected= (22C2 * 14C2) = 21021 ways.