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I've tried both expanding the binomials as well as trying to deduce something from the hypergeometric distribution, but I don't see how to prove:

$${N\choose n}^{-1}\sum_{i\geq j}{M\choose i}{N-M\choose n-i}{i\choose j}\leq {n\choose j}\left(\frac{M}{N}\right)^j.$$

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Using the factorial formulas for binomial coefficients and Vandermonde's Identity, assuming $M\le N$, we get $$ \begin{align} \hspace{-1cm}\binom{N}{n}^{-1}\sum_{i=j}^n\binom{M}{i}\binom{N-M}{n-i}\binom{i}{j} &=\binom{N}{n}^{-1}\sum_{i=j}^n\binom{M}{j}\binom{N-M}{n-i}\binom{M-j}{i-j}\tag{1}\\ &=\binom{N}{n}^{-1}\binom{M}{j}\binom{N-j}{n-j}\tag{2}\\ &=\binom{M}{j}\binom{N}{j}^{-1}\binom{n}{j}\tag{3}\\ &\le\left(\frac MN\right)^j\binom{n}{j}\tag{4} \end{align} $$ Explanation:
$(1)$: $\binom{M}{i}\binom{i}{j} =\frac{M!}{i!\,(M-i)!}\frac{i!}{j!\,(i-j)!} =\frac{M!}{j!\,(M-j)!}\frac{(M-j)!}{(M-i)!\,(i-j)!} =\binom{M}{j}\binom{M-j}{i-j}$
$(2)$: Vandermonde $\implies\sum\limits_{i=j}^n\binom{N-M}{n-i}\binom{M-j}{i-j}=\binom{N-j}{n-j}$
$(3)$: $\binom{N}{n}^{-1}\binom{N-j}{n-j}=\frac{n!\,(N-n)!}{N!}\frac{(N-j)!}{(N-n)!\,(n-j)!}=\frac{n!}{j!\,(n-j)!}\frac{j!\,(N-j)!}{N!}=\binom{N}{j}^{-1}\binom{n}{j}$
$(4)$: $M\le N\implies\frac{M}{N}\frac{M-1}{N-1}\frac{M-2}{N-2}\cdots\frac{M-j+1}{N-j+1}\le\left(\frac MN\right)^j$

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