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A committee of 12 is to be selected from 10 men and 10 women. In how many ways can the selection be carried out if
1 There are no restrictions
2 There must be six men and six women
3 There must be more women than men
4 There must be at least eight men?
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1) If there are no restrictions then you need to choose 12 out of 20. --- 20C12
2) 6 men and 6 women. 12C6 + 12C6
and I dont know how to do to 3 and 4.
Im a Discrete math student myself and we just started to do combinatorics and permutations so maybe my 2 answers are wrong and I'd like to know the answer to the last parts.
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For 2), you want 10C6 instead of 12C6. Also, they're multiplied instead of added.
For 3 and 4, I think you have to work out specific cases and add them all up.
So for 3), you'd work out the number of possible combinations that contain:
- 7 women and 5 men
- 8 women and 4 men
- 9 women and 3 men
- 10 women and 2 men
Then add those four values together to get an answer.
4) can be done similarly.
Why did the vector cross the road?
It wanted to be normal.
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