How to arrange 4 $T$'s, 4 $G$'s and 3 $W$'s if the $W$'s can't be after a $T$? (in a different way) Problem
How many ways can you arrange 4 $T$'s, 4 $G$'s and 3 $W$'s if the $W$'s can't be after a $T$?
(For example, $WWGGWGGTTTT$ is correct, but $WGGTWTGGTWT$ is not.)
It's not really a difficult problem, let 11 spaces, like this:
$$- - - - - - - - - - - $$
Let a $W$ to be a separator between the $T$'s and other letters. ( This letter will not be considered in the arrangement, but it's still being part of $W$'s
You can notice that the last $W$ has to be a least in the third space, then you can arrange the $G$'s and the $T$'s and you get
$$\binom{2}{2} \binom{8}{4}$$
Then, if you move your separator to the fourth space you get
$$\binom{3}{2} \binom{7}{4}$$
And so on until the 7th space. Then you get
$$\binom{2}{2} \binom{8}{4} + \binom{3}{2} \binom{7}{4} +\binom{4}{2} \binom{6}{4} + \binom{5}{2} \binom{5}{4} + \binom{6}{2} \binom{4}{4} $$
And it's the same as 330.
But that's not what i wanted to ask, i'm wondering how to know that $$\binom{11}{7}$$
is the answer (because it is too). I mean, what would you have to do to know it?

If we remove the $G$'s from any valid string, we will get the string $WWWTTTT$.
So choose $7$ elements from the $11$ slots available. From left to right, first put $W$'s in the three leftmost of these; and put the $4$ $T$'s in the remaining four (rightmost) of those slots. Put $G$'s in the other slots.
Alternatively (since ${11 \choose 7} = {11 \choose 4}$), choose the $4$ slots where $G$'s will be placed. Then insert the $W$'s in the remaining slots, starting from the left; then the rest are $T$'s.
