I'm having difficulties with the problem below (and problems that are similar in setup). I stumbled across it in a practice GMAT from MBA.com:

A 4-person task force is to be formed from the 4 men and 3 women who work in Company G's human resources department. If there are to be 2 men and 2 women on this task force, how many different task forces can be formed?

Do I use a factorial to solve this problem?

[The correct answer is 18. I'm struggling to achieve it.] Help would be much appreciated.

This is a classic combination problem..almost as basic as it gets..

Let's say
A: # of ways you can select 2 men out of a group of 4
B: # of ways you can select 2 women out of a group of 3
answer is A*B.

using combination it would (4 C 2)(3 C 2) = (4!/2!2!)(3) = 18

=============================================

If you don't wanna go by the formula, you can solve it using permutations:

Let's calculate for A first:
You have 2 spots to fill and you have 4 items to chose from.
Clearly, you will have 4 options for the 1st and 3 options for the 2nd spot. Therefore 4*3.
Mind you, you need to get rid of redundancies. 4*3 is the number of ways you can pick 2 men from a group of 4 and line them up if the order in which they stood in the line mattered. However, the order doesn't matter here. That is, it doesn't matter if you end up with John in the 1st spot and Dwayne in 2nd or vice versa. Therefore, you need to divide 4*3 by the number of ways 2 items in a line can be rearranged: 2!
so you'll have 4*3/2!=6.

similarly for B..
2 spots, 3 to choose from --> 3*2/2! --> 3
voila!

A*B = 18

hope it helps :)

This is very helpful. Forgive me for asking the following, which probably has a very obvious answer, but I'm pretty unfamiliar with this type of problem. We have --

"Let's say
A: # of ways you can select 2 men out of a group of 4
B: # of ways you can select 2 women out of a group of 3
answer is A*B.

using combination it would (4 C 2)(3 C 2) = (4!/2!2!)(3) = 18"

Could you walk me through (to the point of being painfully literal) how you solved this?

I get that (4 C 2) is supposed to represent the # of ways you can select 2 men out of a group of 4 and (3 C 2) is the # of ways you can select 2 women out of a group of 3. What I don't understand is how you went from that to (4!/2!2!)(3).

Thank you so, so much.

sorry I assumed we know:

(n C n-1) = (n C 1) = n

(n C n-1) = n!/(n-1!)*1! = n
try doing the other one. I'm sure you can prove to yourself that it also yields n ;)

that's why I didn't expand (3 C 2). Just wrote 3 instead.

clear now ?

emily, there are formulas that will do that.

with numbers this small, though, there's no reason to know formulas. if there are only a small number of possibilities, just MAKE A LIST!

if the 4 men are 'a', 'b', 'c', and 'd', then here are the possibilities for choosing two of them:
a, b
a, c
a, d
b, c
b, d
c, d
(six different ways)

if the 3 women are 'x', 'y', and 'z', then here are the possibilities for choosing two of them:
x, y
x, z
y, z
(three different ways)

so that's 6 x 3 = 18 ways in total.

_________________
Being well-dressed gives a feeling of inward tranquillity [that] religion is powerless to bestow.
C.F. Forbes

Ron and Parthian --

Both of you have been very helpful! I dug up the chapter on Combinatorics and have delved into it a bit to gain a better understanding of the concept. Thanks so much for taking the time to respond! It's been much appreciated.

glad to help.

_________________
Being well-dressed gives a feeling of inward tranquillity [that] religion is powerless to bestow.
C.F. Forbes

:)

:-) Thanks, everyone!

_________________
Jamie Nelson
ManhattanGMAT Instructor