I was solving a problem and am stuck with this expression. Any leads on how can I simplify this expression?

$$\frac{{\sum\limits_{x=Q}^{N-P+Q} (x-Q) \binom{x}{Q} \binom{N-x}{P-Q}}}{{\sum\limits_{x=Q}^{N-P+Q} \binom{x}{Q} \binom{N-x}{P-Q}}}$$

UPDATE: I realized a mistake. expression updated.

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    $\begingroup$ For the updated question, note that $(x-Q)\binom xQ=(Q+1)\binom x{Q+1}$, and you can let the sum start at $Q+1$ since the $x=Q$ term is zero. $\endgroup$ – joriki Jan 19 '13 at 13:43
  • $\begingroup$ @joriki Thanks, using that with vandermonde identity simplified the equation. $\endgroup$ – nims Jan 19 '13 at 19:35
  • $\begingroup$ Are you sure? I make it $$ (Q+1)\frac{\binom{N+1}{P+2}}{\binom{N+1}{P+1}}=\frac{(Q+1)(N-P)}{P+2}\;. $$ (This refers to a comment you'd made but now edited, in which you said the result was $(Q+1)/(P+2)$. $\endgroup$ – joriki Jan 19 '13 at 19:39
  • $\begingroup$ @joriki my bad, i was using that expression in another expression where $(N-P)$ was getting cancelled. :) $\endgroup$ – nims Jan 19 '13 at 19:42

There is a variation of the Vandermonde identity that reads, for $k,m,n\in\mathbf N$: $$ \sum_{i=0}^k\binom im\binom{k-i}n=\binom{k+1}{m+n+1}. $$ Here is how you can remember it: let $0\leq a_0<\cdots<a_{m+n}\leq k$ be one of the $\binom{k+1}{m+n+1}$ subsets of $m+n+1$ numbers $a_j$ from the $k+1$-set $\{0,\ldots,k\}$, arranged increasingly. Put $i=a_m$, then there are $\binom im$ choices left for $a_0,\ldots,a_{m-1}$, and $\binom{k-i}n$ choices for $a_{m+1},\ldots,a_{m+n}$.

One can restrict the range of $i$ to the values $m\leq i\leq k-n$, as other terms contribute $0$.

So your expression simplfies to $$ \frac{{\sum\limits_{x=Q}^{N-P+Q} \binom{x-1}{Q} \binom{N-x}{P-Q}}}{{\sum\limits_{x=Q}^{N-P+Q} \binom{x}{Q} \binom{N-x}{P-Q}}}= \frac{\binom{N}{P+1}}{\binom{N+1}{P+1}}=\frac{N-P}{N+1}. $$

  • $\begingroup$ Thanks for the reply. I realized i made a mistake. What to do if the numerator becomes $$\sum\limits_{x=Q}^{N-P+Q} (x-Q) \binom{x}{Q} \binom{N-x}{P-Q}$$ $\endgroup$ – nims Jan 19 '13 at 13:05
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    $\begingroup$ @nims: If you meant another question than the one you posed, please change (edit) the question (and indicate there that you did so). $\endgroup$ – Marc van Leeuwen Jan 19 '13 at 13:41
  • $\begingroup$ Thanks, turns out the new expression can be easily solved by little operation as suggested in the comments and then applying Vandermonde identity as you suggested. $\endgroup$ – nims Jan 19 '13 at 19:40

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