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Nov 2016 p63 q1
2838
A committee of 5 people is to be chosen from 4 men and 6 women. William is one of the 4 men and Mary is one of the 6 women. Find the number of different committees that can be chosen if William and Mary refuse to be on the committee together.
Solution
First, calculate the total number of ways to choose 5 people from 10 (4 men + 6 women):
\(\binom{10}{5} = 252\)
Next, calculate the number of ways William and Mary are both in the committee. If William and Mary are both chosen, we need to choose 3 more people from the remaining 8 people:
\(\binom{8}{3} = 56\)
Therefore, the number of committees where William and Mary are not together is: