+0  
 
+1
321
4
avatar+444 

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

Mr.Owl  Dec 4, 2017
 #1
avatar+578 
0

no integers, but there is 3.259

 #2
avatar+444 
0

There is supposed to be an integer.

Mr.Owl  Dec 4, 2017
 #3
avatar+93038 
+1

C (23, 4)  =      23! / [ (23 - 4)! * 4! ]   =  23! / [ 19! * 4! ]

C(23,5)  =   23! / [ (23 - 5)! * 5! ]  =  23! / { 18! * 5!]

 

So.......we want to solve this

 

23! / [ 19! * 4! ]  + 23! / [ 18! * 5!]  =  24! / [ (24 - k)! * k! ]

 

1 / [ 19! * 24] + 1 / [ 18! * 120 ]  =  24 / [ (24 - k)! * k! ]

 

[ 5   + 19 ]  / [ 19! * 5!]  =  24 / [ (24 - k)! * k! ]

 

Which implies that

 

19! * 5!   =   (24 - k)! * k!

 

Which implies that    k  = 5

 

But  C (24, 5)  =  C(24, 19)

 

So...... 5 + 19  =  24

 

cool cool cool

CPhill  Dec 4, 2017
 #4
avatar+444 
+1

Thank you so much CPhill!!! I had no idea how to solve that, THANKS!!!

Mr.Owl  Dec 4, 2017

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