There are $4$ girls and $3$ boys but there are only $5$ seats. How many ways can you seat the $3$ boys together?
The order of the seat matters, for example: there's the order $B_1$ $B_2$ $B_3$ $G_2$ $G_4$ and there's $B_2$ $B_3$ $B_1$ $G_2$ $G_4$
Here's my answer: There are $3!$ ways to seat the $3$ boys. The $2$ remaining seats are to be occupied by $2$ out of the $4$ girls, so $^4P_2$. So we now have $3! \cdot $ $^4P_2$.
Lastly, there are $3$ ways to make that arrangement,
$1)$ two girls on the left,
$2)$ two girls on the right, and
$3)$ a girl on both ends.
So my final equation is $3! \cdot $ $^4P_2$ $\cdot 3 = 216$
But then again, that was just a guess, I'm not really sure how to get it. So please confirm if my answer is right, and if it's wrong, please tell me how to get it.