The 16 women and 10 men of the Math Club need to form a five person fund-raising
ID: 3126458 • Letter: T
Question
The 16 women and 10 men of the Math Club need to form a five person fund-raising committee How many of these five person committees have more women than men? Three of the Math Club members are Heidi, Michelle, and Zach. If at least one of them, but not all three of them, must serve on the committee, how many possible committees of 5 can be formed? If a committee of 5 is chosen, what is the probability that the committee has more men than women and either Michelle or Heidi, but not both are on the committee?Explanation / Answer
Committee members 5 from 16 women and 10 men
If more women than men
(W,M) = (3,2) (4,1) or (5,0)
Hence no of ways = 16C3(10C2)+16C4(10C1) +16C5
= 25200+18200+4368
= 47668
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Assume the 3 are men we have
No of possible committees = Total committees - no of committees not having Heidi, Michelle and Zach+ All 3 included
= 26C5 - (26-3)C5+ 23C2
= 65780-33649+253= 32384
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More men than women either Michelle, or Heidi
(M,w) = (3,2) or (4,1) or (5,0)
Of these men either Michelle or Heidi should be there
Then 1 person from this 2, and remaining men from remaining 8 men and women from 16
No of ways = 2C1{8C2(16C2) +8C3(16C1)+8C4}
= 2[28(120) + 56(16) + 70}
=8652
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