No. of dots: $k$

No. of slots: $n$

If the dots are placed in every combination within the slots, how many palindromes will there be? The dots cannot be superimposed on each other, which means $k \lt n$.

$$n,k,x,y \in \Bbb N$$

$$\text{No. palindromes} =
\begin{cases}
\displaystyle {n \div 2 \choose \ k \div 2}, & n = 2x, \ k = 2y \\[2ex]
\displaystyle{\lfloor n \div 2 \rfloor\choose k \div 2}, & n = 2x +1, \ k = 2y \\[2ex]
\displaystyle{\lfloor n \div 2 \rfloor \choose \lfloor k \div 2\rfloor}, & n = 2x +1, \ k = 2y + 1 \\[2ex]
0, & n =2x, \ k = 2y +1 
\end{cases}$$

This is what I've got. I'm pretty sure it's correct, but I have no formal proof. I think I have an informal proof:

> 
> For a combination of dots in the slots to make a palindrome, the two halves of the string of slots must be mirror images. That means we're dealing with the halves of $n$ and $k$, as for every arrangement of dots there is on one half of the string, there is only one possible arrangement on the other half (that arrangement being the mirror image). As such, if both $n$ and $k$ are even, the number of palindromes will be equal $n/2 \choose k/2$.
> 
> 
> 

> 
> If $n$ is odd, but $k$ is even, then none of the dots may take the central slot. This is because it would then mean an uneven number of dots $2y -1$ would be distributed over an even number of slots $2x +1 -1$. In such a case, there will always be one more or one less on one of the halves. As such, no dot can be placed in the central slot. This reduced the amount of available slots from $n/2$ to $(n-1)/2$ or just $\lfloor n/2 \rfloor$. That means the number of palindromes in this case is $\lfloor n/2 \rfloor \choose k/2$.
> 
> 
> 

> 
> If both $n$ and $k$ are odd, then a dot **must** occupy the central slot. That removes both a slot and a dot from the previously odd number of slots and dots (respectively), meaning we're left with an even number of dots distributed onto an even number of slots. Thus, the number of palindromes is $\lfloor n/2 \rfloor \choose \lfloor k/2 \rfloor$.
> 
> 
> 

> 
> If $n$ is even but $k$ is odd, then there is no central slot to remove a dot, meaning there are no palindromes.
> 
> 
> 

If this is correct, how can I make it into a formal proof? If it is incorrect, where did I go wrong?

**EDIT:** Forgot to add the rule that the dots cannot overlap. There can never be $3$ dots distributed over $2$ slots.