To solve this problem, we need to consider the different cases where at least one woman is included in the selection of three people.
Case 1: 1 woman and 2 men are chosen.
The number of ways to choose 1 woman from 3 is given by the combination \(\binom{3}{1} = 3\).
The number of ways to choose 2 men from 6 is given by the combination \(\binom{6}{2} = 15\).
Thus, the total number of ways for this case is \(3 \times 15 = 45\).
Case 2: 2 women and 1 man are chosen.
The number of ways to choose 2 women from 3 is given by the combination \(\binom{3}{2} = 3\).
The number of ways to choose 1 man from 6 is given by the combination \(\binom{6}{1} = 6\).
Thus, the total number of ways for this case is \(3 \times 6 = 18\).
Case 3: 3 women are chosen.
The number of ways to choose 3 women from 3 is given by the combination \(\binom{3}{3} = 1\).
Adding all these cases together, the total number of ways to choose at least one woman is \(45 + 18 + 1 = 64\).
Alternatively, calculate the total number of ways to choose any 3 people from 9 (6 men + 3 women) without restriction: \(\binom{9}{3} = 84\).
Subtract the number of ways to choose only men (no women): \(\binom{6}{3} = 20\).
Thus, the number of ways to choose at least one woman is \(84 - 20 = 64\).
