How to prove Vandermonde's Identity: $\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$?

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edited Jul 18, 2016 at 16:42

Caleb Stanford

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Vandermonde's Identity : Summations with binomial coefficients$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}$

How can we prove that

$$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?$$

(Presumptive) Source: Theoretical Exercise 8, Ch 1, A First Course in Probability, 8th ed by Sheldon Ross.
Can someone help me solve this equation? How to prove that $$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?$$

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edited Jul 4, 2015 at 19:32

user940

(Presumptive) Source: Theoretical Exercise 8, Ch 1, A First Course in PrProbability, 8th ed by Sheldon Ross.
Can someone help me solve this equation? How to prove that $$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?$$