Suppose that a department contains 12 men and 19 women. How many different committees of 6 members are possible if the committee must have strictly more women than men?
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Suppose that a department contains 12 men and 19 women. How many different committees of 6 members are possible if the committee must have strictly more women than men?
this means a committee can have:Quote:
Originally Posted by asugirl87
(a) 6 women and 0 men, or
(b) 5 women and 1 man, or
(c) 4 women and 2 men
now find each of those and sum them up. do you know how to find each? (hint: think "combinations")
C(6,0)*C(5,1)*C(4,2) is that what i use?
noQuote:
Originally Posted by asugirl87
for (b) for instance. there are C(19,5) ways to choose 5 women from 19 and C(12,1) ways to choose 1 man from 12. thus, for (b) there are C(19,5)*C(12,1) ways to form a committee with 5 women and 1 man
now do the rest similarly