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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) = $9$

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  • $\begingroup$ nobody said you can't be in both committees $\endgroup$ – steven gregory May 23 '18 at 2:14
  • $\begingroup$ @stevengregory erm Richard said he can only in the bad committee $\endgroup$ – Charlz97 May 23 '18 at 2:27
  • $\begingroup$ Yes, he had to say it because it was a possibility. Hence they choose the members of the second group regardless of their choices for the first group - with the execption of Richard. $\endgroup$ – steven gregory May 23 '18 at 5:00
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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.

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