The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is

Question

The value of $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}$ is?

I tried this: $\binom{50}0\binom{50}1+\binom{50}1\binom{50}2+\dots+\binom{50}{49}\binom{50}{50}=2\binom{50}0\binom{50}1+2\binom{50}1\binom{50}2+\dots+\binom{50}{25}^2$

How can I get that this expression is equal to $\binom{100}{51}$?

Answer 1

$$(1+x)^{50}=\binom{50}{0}+\binom{50}{1}x+\cdots+\binom{50}{49}x^{49}+\binom{50}{50}x^{50}$$

$$(x+1)^{50}=\binom{50}{0}x^{50}+\binom{50}{1}x^{49}+\cdots+\binom{50}{49}x+\binom{50}{50}$$

Multiplying we observe that our required sum is the coefficient of $x^{49}$ in $(1+x)^{100}$

$$S=\binom{100}{49}=\binom{100}{51}$$

Answer 2

You have 100 objects and you want to choose 51 of them. So, divide them into two sets of 50 objects (arbitrarily). Choosing 51 from your original set is the same as choosing (50 from first set) and (1 from second set) plus choosing (49 from first set) and (2 from second set) + ..., and all these are mutually exclusive so you don't have to worry about double counting. Note that you will need to swap some $50 \choose n$ and $50 \choose 50 - n$ to see this explicitly.