+248 Find the sum of all integers $k$ such that $\binom{23}{4} + \binom{23}{5} = \binom{24}{k}$.
+2666 Note that \({24 \choose k} = {24 \choose 24 - k}\)
This means that the sum of all integers \(k\), that work, is \(\color{brown}\boxed{24}\)
+118608 Builderboi, Your answer is correct but i had to think really hard before I could see the relevance of your explanation.
+2666 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 \(n = 24\)
+118608 Yes I get that, and it is good logic, but how did you know that k was defined at all.
OR maybe k is 12 and then there would not be a second k value.
....
I guess it was a reasonable assumtion that k was defined....and that it was not 12.
And yes if any such k exists (which I know it does becasue I worked it out ) and k is not 12, then there will be 2 k values, and their sum is 24 just as you have said. 
