$5$ $A$'s and $6$ $B$'s are arranged in a row. Find the probability that a choosen arrangement is palindrome.
I tried out all the ways a palindrome could be formed. This was favourable outcomes.
All possible outcomes were $\frac{11!}{5!6!}$ Probability was favourable upon total outcomes.
What is the alternate way to find out the number of palindromes? I assume it involves selecting gaps from a row and then arranging the alphabets.
$6$th place should be A and places $1$ to $5$ uniquely determines places $7$ to $11$, so there are $\dfrac{5!}{3!2!}$ palindromes.