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# help

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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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n = 5

8C3 + 8C4 =9C5

56   +  70   = 126

Dec 8, 2018
#2
+99331
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$$\binom{8}{3} + \binom{8}{4} = \binom{9}{n}\\ LHS\\ =\binom{8}{3} + \binom{8}{4} \\ =\frac{8!}{3!\;5!}+\frac{8!}{4!\;4!}\\ =\frac{9! }{3!\;5!*9}+\frac{9!}{4!\;4!*9}\\ =\frac{9! }{3!\;4!*5*9}+\frac{9!}{3!*4\;*4!*9}\\ =\frac{9!}{3!4!*9}\left( \frac{1}{5}+\frac{1}{4} \right)\\ =\frac{9!}{3!4!*9}\left( \frac{9}{20} \right)\\ =\frac{9!}{3!4!}\left( \frac{1}{4*5} \right)\\ =\frac{9!}{4!5!}\\ =\binom{9}{4}\quad or \quad \binom{9}{5}$$

So the largest value of n is 5

Dec 8, 2018
#3
+98172
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Nicely done, Melody !!!!

CPhill  Dec 8, 2018
#4
+99331
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Thanks Chris :)

Melody  Dec 8, 2018
#5
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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
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Thanks Tertre, I figured that there must be but I am not good at remembering tricks. :/

Melody  Dec 8, 2018
#7
+3994
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Welcome! I just learned this a few days ago under Pascal's triangle.

tertre  Dec 8, 2018
#8
+99331
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Of course, how silly of me, I should have thought of that!

Melody  Dec 9, 2018