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

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