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For $n\ge 0$, show $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$.

Over a well ordered set of non-negative integers, induction can be used.

Base case: $n=0$
$\binom{x}{0} \binom{y}{0} = 1.\frac{(y)!}{(y)! \,\,(0)!}=1$.

Taking the rhs, get : $\binom{x+y}{0} = 1$.

Induction hypothesis: it is true for natural number $n$.

So, $\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k} = \binom{x+y}{n}$

Inductive step: need show for next natural $n+1$

$\sum_{k=0}^{n+1} \binom{x}{k} \binom{y}{n-k+1} = \binom{x+y}{n+1}$

This should lead to :

$\binom{x+y}{n} + \binom{x}{n+1}\binom{y}{0}= \binom{x+y}{n+1}$

get on lhs:

$\binom{x+y}{n} + 1.\binom{x}{n+1}$

$\implies \frac{(x+y)!}{n!(x+y-n)!} + \frac{x!}{(n+1)!(x-n-1)!}$

Please help to proceed further, or suggest alternative approach.

jiten
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  • Have you ever heard of a story proof before? – Matthew H. Feb 03 '21 at 04:36
  • @MatthewPilling Never. – jiten Feb 03 '21 at 04:38
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    Suppose a bag contains $x=7$ apples and $y=10$ oranges. In how many ways can you select $n=6$ pieces of fruit from this bag? – Matthew H. Feb 03 '21 at 04:39
  • @MatthewPilling It would be sum of all cases where $n=n_1+n_2$ are from either, with $n_1={0,\cdots, n}, n_2={0,. \cdots, n}$. For $n_1$, all apples are identical, so combinations are taken; same for oranges. As all such combinations are dependent, so form product. Then take sum of all mutually independent cases. Agreed that understand now the logic, but the derivation above is going nowhere using MI. – jiten Feb 03 '21 at 04:43
  • I think the equation you derive from the induction hypothesis is strang. – User Feb 03 '21 at 04:45
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    Is https://math.stackexchange.com/questions/1802387/prove-sum-k-0n-binomnk-binomm-nn-k-binommn?rq=1 related? – kensaii Feb 03 '21 at 04:47
  • @MatthewPilling want to add that $n_1+n_2=n$, that completes the logic needed. – jiten Feb 03 '21 at 04:59
  • @User Please tell where it goes wrong. – jiten Feb 03 '21 at 05:01
  • @kensaii the lhs is having different form (though, the logic is the same). So, MI appln. is difficult here. Or, do you mean should transform this question to that one in order to apply MI. If am correct, then MI appln. needs a particular form. – jiten Feb 03 '21 at 05:11
  • @MatthewPilling please provide some hint on logic for problem:: For $n\ge 0$, show $\sum_{k=0}^{n} \binom{2k}{k} \binom{2(n-k)}{n-k} = 4^n$. It is now not only having variable $k$ (in number of items to be chosen in each category) as variable, but also having the quantity available in each category as variable. I mean, for this post, $x,y$ were fixed before-hand. – jiten Feb 03 '21 at 05:37

2 Answers2

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$$S=\sum_{k=0}^{n} \binom{x}{k} \binom{y}{n-k}$$ $$S=\sum_{k=0}^{n}[t^k]~ (1+t)^x~ [t^{n-k}]~ (1+t)^y$$ $$S=[t^n] (1+t)^{x+y}={x+y \choose n}.$$ Here $[t^j]f(t)$ means co-efficient of $t^j$ in $f(t).$

Z Ahmed
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  • Seems you have transformed the problem to a different form as said by @kensaii. If so, then my comment to kensaii is correct, that MI appln. needs a particular form. – jiten Feb 03 '21 at 05:17
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There are much better proof strategies in my opinion, but here's an inductive proof of the Vandermonde convolution identity.

Joshua P. Swanson
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  • Seems need transformation of the problem to a different form as also stated by @kensaii. If so, then my comment to kensaii is correct, that MI appln. needs a particular form. – jiten Feb 03 '21 at 05:20
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    @jiten No, the linked question uses exactly your form with the variable names changed. In your notation, the induction is on $x+y+n$, say. – Joshua P. Swanson Feb 03 '21 at 06:40