0

Prove that $$\sum_{i=0}^{50} {50 \choose i}^2 = {100 \choose 50}$$

I tried writing the following but did not get anywhere.

$$ {50 \choose 0} {50 \choose50} + {50 \choose 1} {50 \choose49} + \dots + {50 \choose 50} {50 \choose 0} $$

I even thought about pairing the $n$-th term and the respective $51$st term and taking $2$ common (except for $50 \choose 25$, which is alone). I am too scared to take the LCM. Is there any special property that I am unaware of?

To be honest, I don't have any idea from where to even start.

Rodrigo de Azevedo
  • 1
Gojo
  • 13
  • 2
    See this. – TheSilverDoe Jul 14 '23 at 13:02
  • 2
    Consider choosing $50$ objects out of $100$, which is what the right side is counting. Now, fix a split of the $100$ objects into two sets of $50$ each. Consider selecting the $50$ objects by selecting $k$ from the first set and $50-k$ from the second. Note that the number of ways to do that is $\binom{50}{k}\binom{50}{50-k}$. – NDB Jul 14 '23 at 13:03
  • @NDB 's answer is the best (most natural and informative) way to do this. But if you can't see that, induction is always a good approach for these types of question. – Chris Lewis Jul 14 '23 at 13:22
  • @TheSilverDoe and NDB that was helpful, thank you! – Gojo Jul 15 '23 at 12:13

1 Answers1

1

A quick solution by double counting can be given as noted in the above comment. Another standard algebraic approach (which is the same in essence) but it is a cute idea to use polynomials for those kinds of things is the following.

Consider the polynomial:

$$(x+1)^{100}$$

In view of Newton the coefficient of $x^{50}$ when expanding the above polynomial is $\binom{100}{50}$ . On the other hand you can expand it in a different way, one may write:

$$ (x+1)^{100} = (x+1)^{50}(x+1)^{50} $$

And expand after.Then the coefficient of $x^{50}$ by polynomial multiplication and using the binomial must be:

$$ \sum_{i=0}^{50} \binom{50}{i} \binom{50}{50-i} = \sum_{i=0}^{50} \binom{50}{i}^2 $$

but these two coefficients must be the same and the claim follows.

SotArmen
  • 364