The equation for number of ways to put $b$ loops around $b$ of $a$ dots is $$a \choose b$$
How can one modify this so that all palindromes/mirrors are excluded?
This is a continuation of this question. Imagine you have $a$ amount of dots and $b$ amount of loops, where $a>b$. Now imagine you put those loops around the dots, and see how many permutations/combinations you get. Is there an equation to give you the answer if both $a$ and $b$ are known? Here is a picture that shows what I mean:
Here the dots are squares, and the loops are the red squares.
We'll have to work for odd, even $a,b$ separately.
The mirrors can be counted by looking at one half of the string, since the other half is identical.
For a even, b odd, there won't be any mirrors. So following is enough $$\binom{a}{b}$$
For a even, b even, $a=2n$, $b=2m$ there'll be $m$ loops in each half, hence there'll be $\binom{n}{m}$ mirrors
$$\bigcirc \bullet \bullet \boxed{\bullet \bullet \bigcirc}$$
For a odd, b odd, $a=2n+1$, $b=2m+1$ there'll always be a loop in the center. So no. of mirrors $=\binom{n}{m}$
$$\bullet \bigcirc \bullet \boxed{\bigcirc} \boxed{\bullet \bigcirc \bullet } $$
For a odd, b even, $a=2n+1$, $b=2m$ there'll never be a loop on central dot. So no. of mirrors $=\binom{n}{m}$
$$\bigcirc \bigcirc \bullet \boxed{\bullet} \boxed{\bullet \bigcirc \bigcirc } $$
Desired equation is $$\binom{a}{b} - \binom{n}{m}$$ for three of the cases.