Live Q&A Session with Cambridge Admissions Team || Join Chat Room to Attend the Session

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.