"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?"
I went about solving this question by considering 3 cases, first with a single woman and then with 2 and finally with 3.
$$ {3 \choose 1} \ \times \ {6 \choose 2} \ + \ {3 \choose 2} \ \times \ {6 \choose 1} \ + \ {3 \choose 3} $$
Which simplifies to $ \ 45 \ + \ 18 \ + \ 1$ leading to $64$ different choices.
This approach is in fact correct. However, another approach came to my mind as well.
If at least 1 woman must be included in the group then we can simply choose 1 from the 3 women, and fill the remaining 2 slots from the remnant 8 'people'.
$$ {3 \choose 1} \ \times \ {8 \choose 2} $$
This, however, gives the answer $84$. This answer is most certainly wrong but I am unable to explain why the method is incorrect. If someone could explain why this leads to the wrong answer that would be very nice.
from Hot Weekly Questions - Mathematics Stack Exchange
Shaheer ziya
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