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) = $9$

  • $\begingroup$ nobody said you can't be in both committees $\endgroup$ 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. $\endgroup$ May 23 '18 at 5:00

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.