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Hello I guess this equality is true but I don't know how to solve it.

$$\sum_{x=0}^{m(1-\text{sel})} (m-1-x)! (m \cdot \text{sel}) \frac{(m(1-\text{sel}))!}{(m(1-\text{sel})-x)!}(x+1) = \frac{(m+1)!}{1+m \cdot \text{sel}}$$

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    $\begingroup$ What is meaning of $sel$ $\endgroup$ – Adi Dani Apr 5 '13 at 17:45
  • $\begingroup$ @AdiDani: I assumed that $m\cdot sel=n$ and everything works out fine. $\endgroup$ – robjohn Apr 6 '13 at 5:22
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I am assuming that $sel=\frac{n}{m}$, then $$ \begin{align} &\sum_{k=0}^{m-n}(m-1-k)!n\frac{(m-n)!}{(m-n-k)!}(k+1)\\ &=\sum_{k=0}^{m-n}\binom{m-1-k}{n-1}(m-n)!n!(k+1)\\ &=\sum_{k=0}^{m-n}\binom{m-1-k}{n-1}(m-n)!n!((m+1)-(m-k))\\ &=\sum_{k=0}^{m-n}\binom{m-1-k}{n-1}(m+1)(m-n)!n!\\ &-\sum_{k=0}^{m-n}\binom{m-k}{n}n(m-n)!n!\\ &=\binom{m}{n}(m+1)(m-n)!n!\\ &-\binom{m+1}{n+1}n(m-n)!n!\\ &=\binom{m}{n}(m+1)(m-n)!n!\\ &-\binom{m}{n}(m+1)(m-n)!n!\frac{n}{n+1}\\ &=\binom{m}{n}(m+1)(m-n)!n!\frac1{n+1}\\ &=\frac{(m+1)!}{n+1} \end{align} $$

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  • $\begingroup$ Thank you very much. Can you tell me which rules do you use to solve this passage (from 4 to 5) :$$ =\sum_{k=0}^{m-n}\binom{m-1-k}{n-1} = \binom{m}{n}\\ $$ $\endgroup$ – Francesco Apr 6 '13 at 12:48
  • $\begingroup$ @Francesco: The general rule is $$ \sum_{j=m}^{n-k}\binom{n-j}{k}\binom{j}{m}=\binom{n+1}{k+m+1} $$ If we use $m=0$, we get the particular rule I used $$ \sum_{j=0}^{n-k}\binom{n-j}{k}=\binom{n+1}{k+1} $$ $\endgroup$ – robjohn Apr 6 '13 at 14:28
  • $\begingroup$ @robjohn Good job! I didn't think about setting "sel" to a fraction. $\endgroup$ – user940 Apr 6 '13 at 14:38
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If $N$ is a non-negative integer, then the $N$th rising power of $x$ is defined as the following product of $N$ factors: $x^\overline{N}=x(x+1)\cdots(x+N-1)$. For a positive integer $K$, the calculus of rising powers gives $$\sum_{k=1}^K k^\overline{N}={K^\overline{N+1}\over N+1}.$$

Therefore, if $M\leq N$ are non-negative integers $$\begin{eqnarray*} \sum_{x=0}^M {(N-x)!\over (M-x)!}(x+1) &=& \sum_{j=0}^M \sum_{x=j}^M {(N-x)!\over (M-x)!}\\[5pt] &=& \sum_{j=0}^M \sum_{k=1}^{M+1-j} k^\overline{N-M} \\[5pt] &=& {\sum_{j=0}^M (M+1-j)^\overline{N-M+1}\over (N-M+1)} \\[5pt] &=& {\sum_{k=1}^{M+1} k^\overline{N-M+1}\over (N-M+1)} \\[5pt] &=& {(M+1)^\overline{N-M+2}\over (N-M+1)(N-M+2)}.\tag1 \end{eqnarray*}$$


Assume that $j$ and $m$ are positive integers and that "sel" is $j/m$. Writing $s$ for "sel", your expression is $$\sum_{x=0}^{m(1-s)} {(m-1-x)!\over (m(1-s)-x)!}(x+1)\cdot ms\,(m(1-s))! \tag2$$ Setting $N=m-1$ and $M=m(1-s)=m-j$ in (1), we see that (2) reduces to $$ {(m-j+1)^\overline{j+1}\over (ms)(ms+1)}\cdot ms\,(m-j)!={(m+1)!\over ms+1},$$ as you expected.

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