1. **Problem statement:** We need to find the number of ways to select a committee of 5 members from 7 women and 9 men such that at least one woman is on the committee. 2. **Formula and rules:** The total number of ways to select 5 members from 16 people (7 women + 9 men) is given by the combination formula: $$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$ where $n$ is the total number of people and $k$ is the number selected. 3. **Calculate total ways without restriction:** $$\binom{16}{5} = \frac{16!}{5!\times 11!}$$ 4. **Calculate ways with no women (only men):** Selecting all 5 members from 9 men: $$\binom{9}{5} = \frac{9!}{5!\times 4!}$$ 5. **Calculate ways with at least one woman:** This is the total ways minus the ways with no women: $$\binom{16}{5} - \binom{9}{5}$$ 6. **Evaluate the combinations:** $$\binom{16}{5} = 4368$$ $$\binom{9}{5} = 126$$ 7. **Final answer:** $$4368 - 126 = 4242$$ So, there are **4242** ways to select a committee of 5 members with at least one woman.