# data management gr.12

posted by Anonymous

In how many ways can a committee of 8 members be chosen from 7 males and 9 females if the committee must contain at least 3 males?
Please verify if my solution is correct:
total# of 8 members 16C8
#with NO males 7C8
# with at least 3 males
(7C3 x 4C2)+ (7C4 x 4C1)+(7C5 x9C0)

1. MathMate

I would like to see, in your interpretation, what each of the expressions mean.
(7C3 x 4C2)
(7C4 x 4C1)
(7C5 x9C0)

2. Anonymous

(7C3 x4C2) # ways of choosing committee 3 males + 5 females. I made a mistake, I think it should be 9C5 instead of 4C2 => (7C3 x 9C5)
(7C4 x 9C4) choosing 4 males + 4 females (again I made a mistake)
Thanks

3. MathMate

ok, that sounds reasonable.

If you like, you can continue with your method, which is
(7C3*9C5)+(7C4*9C4)+(7C5*9C3)+(7C6*9C2)+(7C7*9C1)

or, equivalently,
16C8-(7C0*9C8+7C1*9C7+7C2*9C6)
which gives (or should give) the same result above.

In doing counting problems, it is always wise to try two different ways of counting the same objective to check the calculations.

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