Arrangement for run of A and B

Question

I found a similiar question

but in that I could understand what do they mean by ways to divide the 6 letters A that are left in 4 parts. E.g. AA,A,,AAA. and after that also few things I cannot undersatnd .

Can anybody make it more clear

Answer 1

I'll explain A) for now

For there to be $4$ runs of $A$ and $4$ runs of be either it will be:

$$(A\dots A)(B\dots B)(A\dots A)\dots$$

or it will be

$$(B\dots B)(A\dots A)(B\dots B)\dots$$

Each of these runs will have at least one of their respective letters in it (specifically not zero) and may have some amount of extra of their respective letters in each. Let $a_1,a_2,a_3,a_4$ be the number of $a$'s in the first, second, third, and fourth runs respectively. Similarly define $b_1,b_2,b_3,b_4$.

We have the following system of equations:

$$\begin{cases} a_1+a_2+a_3+a_4=10\\b_1+b_2+b_3+b_4=8\\1\leq a_i~~\forall i\\ 1\leq b_i~~\forall i\end{cases}$$

A choice of beginning with A's and a choice of the solution $(a_1,a_2,a_3,a_4,b_1,b_2,b_3,b_4)=(3,1,3,3,4,2,1,1)$ for example will correspond to the arrangement $AAABBBBABBAAABAAAB$

The number of solutions to this is the product of the number of solutions to the systems $\begin{cases}a_1+a_2+a_3+a_4=10\\1\leq a_i~~\forall i\end{cases}$ and $\begin{cases}b_1+b_2+b_3+b_4=8\\1\leq b_i~~\forall i\end{cases}$

Solve the number of solutions to each of these sub-problems via stars and bars. For the first system involving $a$'s, there are $\binom{9}{3}$ solutions. To the system with the $b$'s there are $\binom{7}{3}$ solutions.

Choosing whether it is an $A$ at the beginning or a $B$ at the beginning, choosing which solution to the system with $A$'s, then choosing which solution to the system with $B$'s, and applying multiplication principle, we have then a final total of:

$$2\cdot \binom{9}{3}\cdot \binom{7}{3}$$