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I am asked to find the following sum:$$\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}\frac{(k-i)}{M-100}$$

The first two terms in the expression i.e, the binomial coefficients is actually $\binom{M}{100}$ as $\binom{k}{i}$ is the coefficient of $x^i$ in $(1+x)^k$ and $\binom{M-k}{100-i}$ is the coefficient of $x^{100-i}$ in $(1+x)^{M-k}$.

Simplifying things I get an answer which is no was close to the actual answer.

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  • $\begingroup$ What's the relation between $M$ and $k$ ?. $M < k$ ?. $M \geq k$ ?. $M$ and/or $k$ $\in \mathbb{Z}$ or $\mathbb{N}$ ?. $\endgroup$ – Felix Marin Jul 14 '17 at 13:35
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Note that by Vandermonde's identity, $$\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}k=k\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}=k\binom{M}{100}$$ and $$\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}i=\sum_{i=1}^{100}\frac{k}{i}\binom{k-1}{i-1}\binom{M-k}{100-i}i=k\sum_{i=1}^{100}\binom{k-1}{i-1}\binom{M-k}{99-(i-1)}\\ =k\sum_{j=0}^{99}\binom{k-1}{j}\binom{M-k}{99-j}=k\binom{M-1}{99}=\frac{k100}{M}\binom{M}{100}.$$ Hence $$\sum_{i=0}^{100}\binom{k}{i}\binom{M-k}{100-i}\frac{(k-i)}{M-100}= \frac{1}{M-100}\left(k\binom{M}{100}-\frac{k100}{M}\binom{M}{100}\right)= \frac{k}{M}\binom{M}{100}.$$

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  • $\begingroup$ Have edited the question again. Please take a look. $\endgroup$ – Epsilon zero Jul 14 '17 at 11:18
  • $\begingroup$ @Anuran Chowdhury Ok. My hint holds also in this case. $\endgroup$ – Robert Z Jul 14 '17 at 11:20
  • $\begingroup$ Thanks for the help @Robert Z ! I will let you know in a while. $\endgroup$ – Epsilon zero Jul 14 '17 at 11:32
  • $\begingroup$ Have you got (k/m)C(m,100)? I did it but tediously $\endgroup$ – Epsilon zero Jul 14 '17 at 12:00
  • $\begingroup$ @Anuran Chowdhury Yes it is correct. See my edited answer. $\endgroup$ – Robert Z Jul 14 '17 at 12:15

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