Question

how many 5-person committees are possible from a group of 9 people if:
a. there are no restrictions?
b. both Jim and Mary must be on the committee?
c. Either Jim or Mary (but not both) must be on the committee?

Answer 1

45

Answer 2

first one is \(\binom{9}{5}=\frac{9\times 8\times 7\times 6}{4\times 3\times 2}=156\)

Answer 3

Second one is \(\binom 9 3 = 84 \)

Answer 4

Third one is \( 2 \times \binom 8 4 =140\)

Answer 5

I would amend Fool's answer for (b). If Jim and Mary must be on the committee, then you only have 7 other people from which to choose the remaining 3.
so
\[\left(\begin{matrix}7 \\ 3\end{matrix}\right)=35\]

Answer 6

OOoooooooooooooppos!!

Yes, They are already selected so no need to count them in the total number.