Wednesday
March 22, 2017

Post a New Question

Posted by on Saturday, August 30, 2008 at 4:07pm.

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

  • gr12 math - , Sunday, August 31, 2008 at 6:04pm

    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?

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question