hi,

The approach would be like this.

Conditions

1. The team should have only three memebers
2. Team should consist atleast one women.

Therefore the possible combinations are

1. One women, 2 men -> 5c1*6c2 = 75
2. two women, 1 men -> 5c2*6c1 = 60
3. Three women, 0 men -> 5c3 = 10

Therefore the total number of combiantions in 145

Note: The problem in your approach is that after you are selecting one female, you should select only men not from the remaining women and men. (You have exercised the choice of selecting the women)

kaps786 wrote:The English department of a certain college consists of 6 male and 5 female faculty members.

How many ways are there to select a committee of 3 faculty members if atleast one female member must be on the faculty

A 165
B 155
C 145
D 135
E 125


I was doing 5C1*10C2..

ie selecting one female from the 5 first since one female has to be there always..., and then another 2 members could be selected from remaining members 4 females and 6 males...i.e could come from the remaining 10 memebrs

I am doing something wrong if someone can help...

thanks

Brilliant!
Congratulations on making a very important mistake and now you will learn something that lies at the heart of this topic. I always discuss this mistake in the class because it's a very important 'Permutations & Combinations' concept.

Read this post first:
https://www.beatthegmat.com/important-pe ... 11283.html

And now Read on..
I will focus only on your mistake first.
Let the committee members be (M1, M2 ... M6) and (F1, ...F5).

By 5C1, let's say you have selected F1.
Good.

Now, by 10C2, Let's say you select F4 and M3.
Very Good.

We have a team of F1, F4 and M3.
Hmm.

By the same manner of selection let me try to select another team of three.
This time, I'd select F4 first and then select a F1 and M3.

What do I get?
I get F4, F1 and M3.
SAME TEAM! (the order is slightly different, but we're not concerned with the order, are we?)

And this is exactly the mistake that you're making. You are counting some teams again and again unnecessarily. There will be many such repetitions.

Moral of the story: When you select from a group, the females in this case or (A,B,C,D,E) as in post in the link above, Do NOT select members one by one. Select them in one go. If you select one-by-one from a group, you end up getting more teams than required.

Correct way of doing:
Required number of ways = (1F & 2 M) OR (2F & 1M) OR (3F) = 5C1*6C2 + 5C2*6C1 + 5C3 = 75 + 60 +10 = 145

[spoiler](C)[/spoiler] is correct.