gr12 math
posted by milo .
froma group of 6 ladies and 4 men, determine in how many ways a committee of 4 people cxan be selected.
a. with no restirctions
b. 4 women
c. 3 women and 1 men
d. 2 women and 2 men
e. 4 men

With no restrictions, just select any four from ten, which is:
factorial(10) / (factorial(4) x factorial(6))
= (10x9x8x7x6x5x4x3x2x1) / {(6x5x4x3x2x1) x (4x3x2x1)}
= (10x9x8x7) / (4x3x2x1).
If you've got to have four women and two men, then you're choosing four women from six, and also two men from four. The number of ways you can choose four women from six is:
(6x5x4x3x2x1) / {(4x3x2x1) x (2x1)} = (6x5) / (2x1) = 15 [Just to demonstate that, you can easily enumerate them all in this instance, since the number is very small. Call them A, B, C, D, E and F. All the possible combinations are as follows: ABCD, ABCE, ABCF, ABDE, ABDF, ABEF, ACDE, ACDF, ACEF, ADEF, BCDE, BCDF, BCEF, BDEF, CDEF. That's 15 in total.]
The number of ways to choose two men from four is (4x3x2x1) / {(2x1) x (2x1)} = 6. [Again, you can easily enumerate them all here. Call them P, Q, R and S. All the possible combinations are PQ, PR, PS, QR, QS and RS. That's 6 in total.]
The total number of ways of getting BOTH (four women from six) AND (two men from four) is just the product of those two previous answers.
Are you able to tackle the remaining parts of the question now?
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