Find the sum of all integers $k$ such that $\binom{23}{4} + \binom{23}{5} = \binom{24}{k}$.

Note that ${24 \choose k} = {24 \choose 24 - k}$

This means that the sum of all integers $k$, that work, is $\color{brown}\boxed{24}$

Builderboi, Your answer is correct but i had to think really hard before I could see the relevance of your explanation.

Sorry, I didn't do the best job... Hopefully this is better!

There will be some answer for $k$, but there will also be another answer, $24 - k$.

This is because ${n \choose k} = {n! \over k!(n-k)!}$. However, ${n \choose n - k} = {n! \over (n-k)!k!}$. For example, ${5 \choose 2} = {5! \over 2! 3!} = 10$, and likewise, ${5 \choose 3}=10$.

The sum of 2 and 3, which are $k$, and $n - k$ ,respectively, both sum to $n$, or 5. 

Likewise, applying the same logic here, we find that the sum will be $\color{brown}\boxed{24}$, because $n = 24$