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The better approach might be to figure out how many different committees can be formed regardless of the rule that it must contain 1 woman.

That could be \(C_{10}^3 = 120\). Then subtract out the number of committees that do not conform to the rule that at least 1 woman must be on the committee. The committees that do not conform to the rule are the committees that are made up entirely of men. The is represented by \(C_6^3=20\) because we're figuring out how many ways can you select all 3 from the 6 men present. This gives you 120 - 20 = 100, the answer.

abhaypratapsingh wrote:

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 ?

_________________

------------------------------------ J Allen Morris **I'm pretty sure I'm right, but then again, I'm just a guy with his head up his a$$.

Ways to select 3 out of 10 individuals (4W+6M) = 10C3 = 120 Ways to select 3 out of 6 men = 6C3 = 20. These 20 are the combinations that include no women. Every other combination must include atleast 1 woman. Therefore, required ways = 120 - 20 = 100. E _________________

Re: combi ques (m08q04) [#permalink]
28 Oct 2010, 09:38

1

This post received KUDOS

Ok guys, i have a question, and forgive me for it being an amateur one. How do you conclude that out of 10 people, there are 120 combinations of 3? _________________

Re: combi ques (m08q04) [#permalink]
31 Oct 2011, 06:02

SPOT THE PATTERN !!

Thanks for this question.

To add usefully to the thread, I would highlight that you should spot the words "at least one" in the question stem. That should lead you to directly choose the following method: "All possible outcomes - those you don't want", i.e. "all committees - committees with only males".

By doing so, you don't wander trying different things and save some time and calculations. PS: it just took me 7 minutes working the question before realizing that it can be done in a minute... _________________

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 ?

I followed the exact method.. I still can't understand what I did wrong..???

Thanks in advance..

The number you get will have duplications.

Consider the group of three women {ABC} Say you select women A with 4C1, next you can get women B and C with 9C2: {A}{BC}. Now, with the same method you can get the following group {B}{AC}, which is basically the same group.

Re: combi ques (m08q04) [#permalink]
29 Oct 2013, 06:55

What is wrong in this method:

Case I : 1 woman + 2 Men , we have three places to fill _ _ _

I fill first place with a woman, so first place can be filled with 4 ways because there are 4 women, 2nd place can be filled with 6 ways because there are 6 men, third place can be filled with 5 men so in case I total ways : 4 x 6 x 5 = 120

Case II : 2 women + 1 men, three places to fill _ _ _

Lets place 2 women first. so first place can be filled with 4 ways, 2nd place can be filled with 3 ways and 3rd place can be filled with 6 ways (6 men) so

4 X 3 X 6 = 72

Case III : 3 women , three places _ _ _

So 4 X 3 X 2 = 24

So total ways should be : 120 + 72 + 24 = 216

Can anybody please explain what is wrong in this method ?

Re: combi ques (m08q04) [#permalink]
29 Oct 2013, 07:54

dheeraj787 wrote:

What is wrong in this method:

Case I : 1 woman + 2 Men , we have three places to fill _ _ _

I fill first place with a woman, so first place can be filled with 4 ways because there are 4 women, 2nd place can be filled with 6 ways because there are 6 men, third place can be filled with 5 men so in case I total ways : 4 x 6 x 5 = 120

Case II : 2 women + 1 men, three places to fill _ _ _

Lets place 2 women first. so first place can be filled with 4 ways, 2nd place can be filled with 3 ways and 3rd place can be filled with 6 ways (6 men) so

4 X 3 X 6 = 72

Case III : 3 women , three places _ _ _

So 4 X 3 X 2 = 24

So total ways should be : 120 + 72 + 24 = 216

Can anybody please explain what is wrong in this method ?

dude you need not use permutations here, you just have you make combinations. 1. 1 woman and 2 men - 4c1x6c2= 60 2. 2 women and 1 men - 4c2x6c1= 36. 3. all three women- 4c3= 4. Add em all to get 100 cases

Re: combi ques (m08q04) [#permalink]
29 Oct 2013, 09:12

lepatron wrote:

SPOT THE PATTERN !!

Thanks for this question.

To add usefully to the thread, I would highlight that you should spot the words "at least one" in the question stem. That should lead you to directly choose the following method: "All possible outcomes - those you don't want", i.e. "all committees - committees with only males".

By doing so, you don't wander trying different things and save some time and calculations. PS: it just took me 7 minutes working the question before realizing that it can be done in a minute...

Great tip, recognizing that you should subtract combinations with all men instead of adding combinations with women is key.