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Let n and k be positive integers such that \(n<10^6\) and \(\binom{13}{13} + \binom{14}{13} + \binom{15}{13} + \dots + \binom{52}{13} + \binom{53}{13} + \binom{54}{13} = \binom{n}{k}\)
What is the ordered pair (n, k)?

 Feb 25, 2020
 #1
avatar+6179 
+1

\(\text{Using the hockey stick identity}\\ \sum \limits_{n=13}^{54}\dbinom{n}{13} = \dbinom{54+1}{13+1} = \dbinom{55}{14}\)

 

https://en.wikipedia.org/wiki/Hockey-stick_identity

 Feb 25, 2020
 #2
avatar+166 
+1

oh that makes sense to use the hockey stick identity. Thanks for the help!!

mathmathj28  Feb 25, 2020
 #3
avatar+288 
0

That's not an ordered pair though.

AnimalMaster  Feb 27, 2020
 #4
avatar+166 
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The ordered pair would be (55, 14)

mathmathj28  Mar 4, 2020

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