combination

[blaster]

if 4 women and 6 men work in the accounting department in how many ways can a committee of 3 be formed if it has to include at least one woman?

36
60
72
80
100

i can find solution by reverse method (total combination - committee entirely men ) but can't get right answer by direct method (one woman 2 men, 2 woman 1 men etc.)

[Krk]

Here is my approach for this problem.
A bruteforce method:
The condition is "Atleast 1 woman"
There can be three different ways
1. W, M, M
2. W, W, M
3. W, W, W
Combinations for 1. W, M, M:
4C1 * 6C2 = 60

Combinations for 2. W, W, M:
4C2 * 6C1 = 36

Combinations for 3. W, W, W:
4C3 * 6C0 = 4

So adding all the combinations 60+36+4 = 100
Hopefully correct answer is 100

[blaster]

yeap,correct answer is 100.
my mistake was that i just added values,instead of multiply.

[Brent@GMATPrepNow]

if 4 women and 6 men work in the accounting department in how many ways can a committee of 3 be formed if it has to include at least one woman?

36
60
72
80
100

For the benefit of other readers, here's the approach that blaster mentioned:

# of committees with at least 1 woman = total # of committees possible - # of committees with 0 women

Total # of committees possible (i.e., ignoring the restriction) = 10C3 = 120 (choose 3 committee members from the 10 people)

# of committees with 0 women = 6C3 = 20 (choose 3 committee members from the 6 men)

So, # of committees with at least 1 woman = 120 - 20 = 100

Cheers,
Brent

[Brent@GMATPrepNow]

By the way, if you'd like to learn how to quickly calculate combinations (such as 6C3) in your head, you can watch video #17 at https://www.gmatprepnow.com/module/gmat-counting (it's free)

Cheers,
Brent