Simplification of $\binom{50}{0}\binom{50}{1}+\binom{50}{1}\binom{50}{2}+\cdots+\binom{50}{49}\binom{50}{50}$

Question

$$\binom{50}{0}\binom{50}{1}+\binom{50}{1}\binom{50}{2}+\cdots+\binom{50}{49}\binom{50}{50}$$

The above sequence amounts to one of the following options:

1. $\binom{100}{50}$
2. $\binom{100}{51}$
3. $\binom{50}{25}$
4. $\binom{50}{25}^2$

Answer 1

Hint. Suppose you have a sheet with 50 drawings of boys and 50 drawings of girls. You're supposed to color all the boys blue and all the girls pink. However, enraged at the inherent sexism of the exercise, you decide to color exactly 51 of the drawings in the wrong color.

How many ways are there to carry out your protest?

Well, first you choose how many pink girls there are going to be; then compute how many pink boys you need to make 51 wrongs in total, and finally choose which of the drawings are going to be pink.

Answer 2

$$\binom{50}{0}\binom{50}{1}+\binom{50}{1}\binom{50}{2}+\cdots+\binom{50}{49}\binom{50}{50}$$ Using $\binom{n}{r}=\binom{n}{n-r}$ , $$= \binom{50}{0}\binom{50}{49}+\binom{50}{1}\binom{50}{48}+\cdots+\binom{50}{49}\binom{50}{0}$$

Using Vandermonde's identity:

$$=\binom{50+50}{49}=\binom{100}{49}=\binom{100}{51}$$ Option b is correct.