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?

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?

anonymous · Answer

if there are no restrictions, then you are being asked to compute the number of ways you can choose 5 from a set of 9, called "nine choose 5" and written either as 
$$\binom{9}{5}$$ or 
$$_9C_5$$ and computed via 
$$\binom{9}{5}=\frac{9\times 8\times 7\times 6}{4\times 3\times 2}$$ cancel first, multiply last