In this Youtube video, Art of Problem Solving's Richard Rusczyk discusses Pascal's Identity. He begins by giving an intuitive understanding using a concrete example.
In his example, a committee of 3 needs to be formed from a group of 12 (which includes Richard). But that's more: two types of committees need to be formed - one that he calls a "bad" committee, which he must be in, and one that he calls a "good" committee, which he must not be in.
At 2:05, he has formed an equation namely:
Total no. of committees = total possible no of bad committees w/him + total possible number of good committees w/o him i.e.
${12 \choose 3}$ = ${11 \choose 2}$ + ${11 \choose 3}$
This is my first time trying to understand Pascal's Identity and my knowledge of combinatorics is not strong. I would like to ask why the total possible number of good committees is not ${9 \choose 3}$. I am assuming that the good committee is chosen after the bad committee. Therefore, the total pool will have shrunk from $12 - 1$(him) $- 2$(the people in the bad committee) = $9$
The wording is a little misleading. You are forming a single committee of $3$ (not two committees of $3$ each). Each possible committee of $3$ is either "good" or "bad." So the total number of committees is the number of "good" committees plus the number of "bad" committees.