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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
avatar+1747 
+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
avatar+117494 
+1

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

 May 18, 2022
 #3
avatar+1747 
+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
avatar+117494 
0

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.    laugh

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

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

astm48  May 18, 2022

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