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Combinatorial proof for $n > k > 0$:

$$\sum_{j=0}^n {{k}\choose{j}} {{n-k}\choose{j}} = {{n}\choose{k}}$$

I don't get why would that work. On the RHS I just choose $k$ elements from $n$ avaliable. I know that $n > k$ so thats valid.

On the LHS I choose $j$ elements from $k$ avaliable and that j $\in [0, n]$, but $n>k$ so at some point that doesn't make sense.

Also the whole meaning of LHS: I choose $j$ elements from $k$ avaliable, then $j$ elements from $n- k$ avaliable (because $k$ are already chosen). So I end up with $2 \times j$ elements chosen from $n$ avaliable separated by some "inner group border" so to say (namely, $k$).

Does anybody understand any of that? Thanks in advance!

RobPratt
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thefool
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  • Hint: pick small n, and work out an example. Try and observe the pattern – Alborz Aug 09 '23 at 16:26
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    It'll help to see the first coefficient as $\binom{k}{k - j}$, because then you can better see why $k$ elements have been chosen among $n$ elements for the RHS. – Bruno B Aug 09 '23 at 16:38
  • Are you sure the question is correct ? when j gets bigger than k, $\binom{k}{j}$ equals zero because you're choosing a big number from a small one. So you can change the question to :$$\sum_{j=0}^k {{k}\choose{j}} {{n-k}\choose{j}} = {{n}\choose{k}}$$ – madfd adfd Aug 09 '23 at 17:11
  • @BrunoB I get it now. Thank you! – thefool Aug 09 '23 at 20:02

1 Answers1

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In the name of God
Let me explain you with an example. Let n = 5, k =2. For example you have five letters A,B,C,D,E and you want to choose two letters from them . One way is choosing 2 from 5. (RHS)
Another way, is dividing these five letters into two groups randomly.( in a way that first group has n-k elements and second group has k elements).
For example A,B,C is the first group and D,E is the second group. By choosing n letters from first group and n letters from the second group, you choose the letters that are going to relocate with each other. when you choose 0 from first group and 0 from second group, the groups remain unchanged. when you choose 1 from first group and 1 from second group, there are $\binom{3}{1}$ $\binom{2}{1}$ = 6 ways that the groups can exchange one of their letters, and so on.
And of course, the second group that has k elements gives you the combinations. which equals LHS.

I hope you've understood. Algebraic proof should be easier, BTW.

madfd adfd
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  • I don't understand how on the LHS enumerating all the ways in which 2 groups ($k$ and $n-k$) can exchange letters would be equal to RHS, which is simply choosing $k$ elements from a set of n elements. @madfd adfd – thefool Aug 09 '23 at 19:41
  • Because number of all the ways that second group can exchange letters with first group is equal to number of all possible forms of second group. In the example that I said, all the possible combinations have three states compared to D,E : They have 0 letters in common. They have 1 letter in common and they have 2 letters in common. There is 1 way that second group remains unchanged, there are 6 ways that it can exchange one of its letters, and there are 3 ways that it can exchange all of its letters. 1 + 6 + 3 is equal to $\binom{5}{2}$ – madfd adfd Aug 09 '23 at 21:27