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Nov 2006 p6 q6
2877
Six men and three women are standing in a supermarket queue.
Three of the people in the queue are chosen to take part in a customer survey. How many different choices are possible if at least one woman must be included?
Solution
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 \(84 - 20 = 64\).
