# Is Fred on the committee?

I was just helping a student with his AoPS homework, when I came across the following related problems:

Eight people, including Fred, are in a club.  They decide to form a 3 person committee.  How many possible committees can be formed?

So we are choosing 3 people out of 8 or ${8 \choose 3}$.

How many possible committees include Fred?

Since Fred is taking one committee seat, that means we need to choose 2 more people from the remaining seven, or ${7 \choose 2}$.

How many possible committees do not include Fred?

Since we can’t choose Fred, we need to choose 3 members from the remaining 7 or ${7 \choose 3}$.

Since the total number of committees is equal to the number of of committees with Fred plus the number of committees without Fred, then we can say

${8 \choose 3} = {7 \choose 2} +{7 \choose 3}$.

Generally we call this Pascal’s Rule: ${n \choose k} = {n-1 \choose k-1} +{n-1 \choose k}$.