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Difference between revisions of "MIE 2015/Day 1/Problem 7"

(Created page with "==Problem 7== Compute <math>\binom{2016}{5}+\binom{2017}{5}+\binom{2018}{5}+\binom{2019}{5}+\binom{2020}{5}+\binom{2016}{6}</math> (a) <math>\binom{2020}{6}</math> (b) <...")
 
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Latest revision as of 16:18, 11 January 2018

Problem 7

Compute


$\binom{2016}{5}+\binom{2017}{5}+\binom{2018}{5}+\binom{2019}{5}+\binom{2020}{5}+\binom{2016}{6}$


(a) $\binom{2020}{6}$


(b) $\binom{2020}{7}$


(c) $\binom{2021}{5}$


(d) $\binom{2021}{6}$


(e) $\binom{2022}{5}$


Solution

Remember that $\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}{k}$ for $n\geq1$ and $0\leq k\leq n$.


Thus,


$\binom{2016}{5}+\binom{2017}{5}+\binom{2018}{5}+\binom{2019}{5}+\binom{2020}{5}+\binom{2016}{6}$


$\underbrace{\underbrace{\underbrace{\underbrace{\underbrace{\binom{2016}{6}+\binom{2016}{5}}_{\binom{2017}{6}}+\binom{2017}{5}}_{\binom{2018}{6}}+\binom{2018}{5}}_{\binom{2019}{6}}+\binom{2019}{5}}_{\binom{2020}{6}}+\binom{2020}{5}}_{\binom{2021}{6}}$


$\boxed{\text{D}}$

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