How to arrange letters in blanks where letters are repeating aand satisfying the given condition? 7blank spaces arranged in a row. We have 2 A's, 3B's & 2 C's which are to be filled in these seven blanks such that the same letter never comes in consecutive blank space.In how many ways can this be done if B is always filled in the first blank?
My soln:
First blank 1way i.e B
2nd blank: 2ways i.e A/C
3nd blank: 4ways i.e A-b/c / C-a/b
Looks like binary tree at this stage so the 7th place will have that many no of nodes as the possible string arrangements.
No of nodes in 7th place = 2^6 =64
Is it right as the multiple choice didn't have this option?
 A: Starting with $BA$ gives the same number of possibilities as starting with $BC$. So it is enough to start with $BA$ and double the number of outcomes. That gives $2\times10=20$ possibilities. The endresults are embraced with braces $[$ and $]$.
$\begin{array}{cccccc}
 &  &  & \left[BABACBC\right]\\
 &  &  &  &  & \left[BABCABC\right]\\
 &  &  &  & BABCA\\
 &  & BAB &  &  & \left[BABCACB\right]\\
 &  &  & BABC\\
 &  &  &  &  & \left[BABCBAC\right]\\
 &  &  &  & BABCB\\
 & BA &  &  &  & \left[BABCBCA\right]\\
\\ &  &  & \left[BACABCB\right]\\
\\ &  &  &  &  & \left[BACBABC\right]\\
 &  & BAC &  & BACBA\\
 &  &  &  &  & \left[BACBACB\right]\\
 &  &  & BACB\\
 &  &  &  &  & \left[BACBCAB\right]\\
 &  &  &  & BACBC\\
B &  &  &  &  & \left[BACBCBA\right]\end{array}$
A: Another way would be:
(a) Forget about starting B. We now have 2A's, 2B's, 2C's.
(b) Next letter can't be B, so count  $\frac23$ of permissible arrangements using inclusion-exclusion.
[When 2A's are together, e.g., treat as one letter, so 5 objects with B's and C's repeated] 
$\frac23\left[\frac{6!}{2!2!2!} - 3\cdot\frac{5!}{2!2!} + 3\cdot\frac{4!}{2!} -3!\right] = 20$
