Problem Solving

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
select 1 women 4C1 and then selecting 2 from rest of 9 is 9C2 , so total= 4C1 + 9C2

But the answer is 100 .
What Am I doing wrong ?

can someone pls help to explain?
thnks,

This is a tough question that I definitely would not have been able to do in 2 minutes, but I think I got it. I am no expert, but here is an attempt to explain the question.

You need to do several separate combinations:

The possible groupings will be 3W, 2W/1M, 1W/2M (W=women, M=man)

For 3W: 3 combination 4 = 4

For 2W/1M: There are 2C4 possibilities for 2W = 6. Each of those unique pairs of 2W can be paired with one of the 6 men: 6*6=36

For 1W/2M: There are 4 options for 1 woman and 2C6 possibilities for 2 men. 2C6=15. 15*4 = 60

So, adding 4 + 36 + 60 = 100

A question like that would really throw me off on timing. Would love to know if there is an easier way to do this question.

easier way :
Total no ways = 10C3 =120
total no ways includig only men = 6C3 = 20

total no ways including at least 1 woman = 120 - 20 = 100.

Can someone explain to me what exactly 10c3 and 6c3, or any similar expression for that matter is? I'm just now starting to understand permutation/combination questions and have no idea what XcX/XcX (10C4) means. Thanks in advance...

at least one women can b chosen in following 3 ways
1w+2m; 2w+1m; 3w
this can be done as
4c1*6c2+4c2*6c1+4c3
60+36+4
=100

moved to the PS section