Problem 6.1.20 Part B From The Arts and Crafts of Problem Solving

Question

Could someone help me with part B of problem 6.1.20? I had attached the solution to part A so you guys (and gals) could get the gist of how the want us to reason the identity. Someone told me that it is very similar to Part A. They mentioned to consider selecting n+1 people from a group of 2n+2 people where there are 2n boys and 2 girls. I've considered and thought...

There are three ways this could go. You could choose all boys, all girls and the whatever is left of n+1 out of all the boys, or one girl and n out of all boys. The first part of the identity would then make sense, all boys would be the binomial theorem of n+1 out of 2n. But I don't get the rest. Thanks in advance.

The Problem - 6.1.20 Part B

The Solution to 6.1.20 Part A

Answer 1

Hint: the second term is the number of ways to choose $n+1$ people such that exactly one is a girl, and the third term is the number of ways to choose $n+1$ people such that both girls are chosen.

Second term: there are $\binom{2n}{n}$ ways to choose $n$ boys, and $2$ ways to choose a girl.

$~$

Third term: there are $\binom{2n}{n-1}$ ways to choose $n-1$ boys, and $1$ way to choose both girls.