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What is the largest integer \(n\) for which \(\binom{8}{3} + \binom{8}{4} = \binom{9}{n}\)?

 Dec 8, 2018
 #1
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+2

n = 5

8C3 + 8C4 =9C5

56   +  70   = 126 

 Dec 8, 2018
 #2
avatar+99331 
+2

\(\binom{8}{3} + \binom{8}{4} = \binom{9}{n}\\ LHS\\ =\binom{8}{3} + \binom{8}{4} \\ =\frac{8!}{3!\;5!}+\frac{8!}{4!\;4!}\\ =\frac{9! }{3!\;5!*9}+\frac{9!}{4!\;4!*9}\\ =\frac{9! }{3!\;4!*5*9}+\frac{9!}{3!*4\;*4!*9}\\ =\frac{9!}{3!4!*9}\left( \frac{1}{5}+\frac{1}{4} \right)\\ =\frac{9!}{3!4!*9}\left( \frac{9}{20} \right)\\ =\frac{9!}{3!4!}\left( \frac{1}{4*5} \right)\\ =\frac{9!}{4!5!}\\ =\binom{9}{4}\quad or \quad \binom{9}{5}\)

 

So the largest value of n is 5

 Dec 8, 2018
 #3
avatar+98172 
+2

Nicely done, Melody !!!!

 

 

cool cool cool

CPhill  Dec 8, 2018
 #4
avatar+99331 
+1

Thanks Chris :)

Melody  Dec 8, 2018
 #5
avatar+3994 
+2

There is actually a trick for this! \( {n \choose k} \)+\( {n \choose k+1} \)=\( {n+1 \choose k+1} \), so in this case, we have \( {8 \choose 3} + {8 \choose 4} = {9 \choose 4} \) or \( {9 \choose 5} \). Thus, \(\boxed{n=5}.\)

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 Dec 8, 2018
 #6
avatar+99331 
+1

Thanks Tertre, I figured that there must be but I am not good at remembering tricks. :/

Melody  Dec 8, 2018
 #7
avatar+3994 
+1

Welcome! I just learned this a few days ago under Pascal's triangle.

tertre  Dec 8, 2018
 #8
avatar+99331 
0

Of course, how silly of me, I should have thought of that!

Melody  Dec 9, 2018

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