Posted by bballer on Wednesday, September 26, 2012 at 12:30am.
Suppose that a department contains 11 men and 19 women. How many ways are there to form a committee with 6 members if it must have strictly more women than men?

math  Reiny, Wednesday, September 26, 2012 at 12:44am
cases:
6W0M
5W1M
4W2M
3W3M  no good
number of cases
= C(19,6) x C(11,0) + C(19,5) x C(11,1) + C(19,4) x C(11,2)
= 27132 + 127908 + 213180
= 368220
Answer This Question
Related Questions
 Discreate Math  Suppose that a department contains 13 men and 19 women. How ...
 discrete math  Suppose that a department contains 10 men and 17 women. How many...
 discrete math  Suppose that a department contains 10 men and 17 women. How many...
 statistics  Suppose that a department contains 8 men and 20 women. How many ...
 math  A department contains 13 men and 20 women. How many ways are there to ...
 math  A department contains 12 men and 17 women. How many ways are there to ...
 business math  out of 7 men and 5 women, 5 members of a committee are selected...
 gr12 math  froma group of 6 ladies and 4 men, determine in how many ways a ...
 math  from a group of women and 4 men, determine in how many ways a committee ...
 statistics permutations and combinations  Selecting a committee: There are 7 ...
More Related Questions