Problem (Combinatorics):
$\quad$ Let $T=2$. In how many ways can $T$ boys and $T+1$ girls be arranged in a row if all the girls must be standing next to each other?

Solution:
First choose the position of the first girl, starting from the left. There are $T+1$ possible positions, and then the positions for the girls are all determined. There are $(T+1)$ ! ways to arrange the girls, and there are $T$ ! ways to arrange the boys, for a total of $(T+1) \cdot(T+1) ! \cdot T !=$ $((T+1) !)^{2}$ arrangements. With $T=2$, the answer is $\mathbf{3 6}$.

Final answer: 36