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Essentially, prove the following summation:

$$\sum_{i=0}^{2x} {{2x \choose 2i}{2i \choose i}{2x-2i \choose x-i}} = {2x \choose x}^2$$

I came across this while attempting question 6 of this paper (BMO1 2019) and while I'm sure there are easier solutions to the problem, this summation seemed interesting in itself. Very similar to this summation is the following:

$$\sum_{i=0}^{x} {{2i \choose i}{2x-2i \choose x-i}} = {2}^{2x}$$

Which I also do not have a proof for. The second one can also be thought of as going down the middle of Pascal's triangle and adding the products of opposite ends of the triangle.

Proofs of either summation, ideally using mathematical ideas which the BMO1 is based around (so basically pre-calculus), would be much appreciated.

Aditya Gupta
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1 Answers1

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This is combinatorial proof. You have $2x$ people whom you will divide into four groups; A, B, C, and D. A and B has equal number of members and C and D has equal number of members.

The first way to do it:

  1. Choose an even number of people who will join either A or B. $\binom{2x}{2i}$
  2. Half of these $2i$ people join A while the rest join B. $\binom{2i}{i}$
  3. Half of $2x-2i$ remaining people join C while the rest join D. $\binom{2x-2i}{x-i}$

Total ways of forming such four groups is $\sum_{i=0}^{x}{\binom{2x}{2i}\binom{2i}{i}\binom{2x-2i}{x-i}}$

The second way to do it:

  1. Choose $x$ people. $\binom{2x}{x}$
  2. Choose $x$ people again, does not have to be different people. $\binom{2x}{x}$
  3. People who got chosen twice join A
  4. People who never got chosen join B
  5. People who got chosen only in first selection join C
  6. People who got chosen only in second selection join D

Total ways of forming such four groups is $\binom{2x}{x}\binom{2x}{x}$

acat3
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