# Tag Info

### Trying to prove equivalence of combinatorial formula and nested summations

The nested summation counts integer tuples $(x_1,x_2,\dots,x_r)$ such that $1 \le x_1 \le x_2 \le \dots \le x_r \le n-r+1$. Performing a change of variables \begin{align} y_1 &= x_1 - 1 \\ y_2 &...
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1 vote

### $(1-x)^{n+a} \sum_{j=0}^\infty \binom{n+j-1}{j}\binom{n+j}{a} x^j = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j$

We need to show $$(1-x)^{n+a} \sum_{m=0}^\infty \binom{n+m-1}{m}\binom{n+m}{a} x^m = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j.$$ Instead, let's move $(1-x)^{n+a}$ to the RHS using a standard ...
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1 vote

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### Limit of $\sum_{j=0}^n (1-j){n\choose j} \frac{1}{(n-1)^j}\left( \frac{n-j}{n} \right)^k$ as $n\to \infty$ (and $k=n$)

I just realised: doesn't this follow from the Dominated Convergence Theorem? We can put the indicator $\mathbb{1}_{j<n}$ in there and let the sum go all the way to infinity (or actually the ...
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Accepted

### How to apply Vandermonde's Identity regarding summation bounds ? $\sum_{t=0}^n\binom{-k-1}{t-k}\binom{-j-1}{n-t-j}=\binom{-k-j-2}{n-j-k}$

Recall that $$\binom{v}{m} = 0 \quad m>v \; \textrm{ or} \; m<0$$ Note that in $$S=\sum_{t=0}^n\binom{-k-1}{t-k}\binom{-j-1}{n-t-j}$$ If $$t<k \implies \displaystyle \binom{-k-1}{t-k}=0$$ ...
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### A game requires 2 players opposite 2 other players, with 6 people available, how many distinct games can take place?

A more pedestrian approach: you need to select 4 people out of 6, for the first one you have six possibilities, second: five etc. But then you need to rule out some permutations: one within each pair, ...
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### A game requires 2 players opposite 2 other players, with 6 people available, how many distinct games can take place?

With the tennis season heating up, we can look at it like arranging doubles tennis matches. $4$ individuals can be selected in $\binom64 = 15$ ways and the tallest among them can be paired with any ...
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### Prove the following combinatorics equality

Here we have Chu-Vandermonde's identity in disguise. We obtain \begin{align*} \color{blue}{\sum_{j=k+1}^{n-m}}&\color{blue}{\binom{j-1}{k}\binom{n-j}{m}}\\ &=\sum_{j=0}^{n-m-k-1}\binom{j+k}{k}...
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1 vote

### Seeking Reference for Identity Involving Binomial Coefficients

You won't find any reference, the claim is not true. $$1\leq 2\\ 1+3\leq 2+2\\$$ but $$3=\binom{1}{2}+\binom{3}{2}\not\leq \binom{2}{2}+\binom{2}{2}=2.$$
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