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:

\(252 - 56 = 196\)