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

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

May 17, 2022

#1
+2448
+1

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

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

May 17, 2022
#2
+118141
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May 18, 2022
#3
+2448
+1

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 $$n = 24$$

May 18, 2022
#4
+118141
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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.

Melody  May 18, 2022
edited by Melody  May 18, 2022
edited by Melody  May 18, 2022
#5
+2
+1

Thank you. What's a great way. I couldn't think of this way.    cupcake 2048

astm48  May 18, 2022