Go to GMAT Club Tests http://gmatclub.com/tests

abhaypratapsingh · **Posted:** Fri Aug 01, 2008 7:01 am


		00:00



	GMAT Timer...

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?

(A) 36
(B) 60
(C) 72
(D) 80
(E) 100

[Reveal] Spoiler: OA

E

Source: GMAT Club Tests - hardest GMAT questions

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 ?

rmangal · **Posted:** Wed Oct 27, 2010 2:48 pm

The answer is E.

Permutations and Combinations is very confusing so I solved this question in this way:
Atleast one women gives us 3 scenarios,

1) All 3 women
2) 1 women and 2 men
3) 2 women and 1 man.

So for all 3 women we get 4 possible ways.
For 1 women and 2 men we get 60 possible ways.
FOr 2 women and 1 man we get 36 possible ways.

So the total is 4+60+36 = 100.

jallenmorris · **Posted:** Fri Aug 01, 2008 7:05 am

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 ?

vaibhavtripathi · **Posted:** Wed Oct 27, 2010 6:17 am

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

durgesh79 · **Posted:** Fri Aug 01, 2008 7:29 am

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 ?

the question asks "at least one Woman" so you can have

1W + 2M = 4C1 * 6C2 = 4 * 15 = 60
2W + 1M = 4C2 * 6C1 = 6 * 6 = 36
3W = 4C3 = 4

Total = 60+36+4 = 100

abhaypratapsingh · **Posted:** Fri Aug 01, 2008 7:31 am

thanks guys..
But what exactly am I doing wrong with my method ?

nmohindru · **Posted:** Fri Aug 01, 2008 9:32 am

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 ?

It says atleast 1 woman

Total no. of ways = 3 women OR 2women + 1man OR 1woman + 2 men

= 4C3 + 4C2*6C1 + 4C1*6C2 = 4 + 36 + 60 = 100

abhaypratapsingh · **Posted:** Fri Aug 01, 2008 9:46 am

i understand "Total no. of ways = 3 women OR 2women + 1man OR 1woman + 2 men"
but I donot understand why following is wrong :

Total = (no. of ways to select one woman ) * (select any two from the rest 9)
= 4C1 * 9C2 = 154

nmohindru · **Posted:** Fri Aug 01, 2008 10:04 am

abhaypratapsingh wrote:

i understand "Total no. of ways = 3 women OR 2women + 1man OR 1woman + 2 men"
but I donot understand why following is wrong :

Total = (no. of ways to select one woman ) * (select any two from the rest 9)
= 4C1 * 9C2 = 154

Because in your method you are double counting

Suppose there were four women A, B, C, D and six men 1,2,3,4,5,6

Then according to you if A is selected then committee can be A,B,C as one of option

If B is selected using 4C1 then B,A,C can be one of option which is same as above.

Similarly A, B , 1 and B, A, 1 will be double count

Therefore if you use your method you have to subtract total number of double counts

tiruraju · **Posted:** Wed Oct 27, 2010 7:37 am

Easy E...10c3 - 6c3

MisterEko · **Posted:** Thu Oct 28, 2010 10:38 am

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?

jallenmorris · **Posted:** Thu Oct 28, 2010 10:55 am

MisterEko wrote:

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?

Don't feel bad about asking that question. There no stupid questions, just stupidly phrased questions. Your question is neither.

The formula is \frac{10!}{3!(10-3)!}

This means on top 10*9*8*7*6*5*4*3*2*1 divided by 3*2*1*7*6*5*4*3*2*1. You can do some cancelling out.

The top 7 through 1 cancels out with the bottom 7 through 1 leaving

\frac{10*9*8}{3*2}

10 / 2 = 5, and 9 /3 = 3, so now we don't have a denominator, and we are left with 5*3*8 = 15 * 8 = 120.

gaurav28jan · **Posted:** Fri Oct 29, 2010 3:12 am

10C3-6C3..this is simplest way of doing the problem...

Answer : E

lepatron · **Posted:** Mon Oct 31, 2011 7:02 am

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...

g106 · **Posted:** Mon Oct 31, 2011 11:36 am

vaibhavtripathi wrote:

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

smart work

ans E

Posted from my mobile device

nishtil · **Posted:** Tue Nov 01, 2011 12:23 am

This is very easy 10c3 -6c3 = 120-20 =100

Jdam · **Posted:** Thu Nov 03, 2011 4:48 pm

jallenmorris wrote:

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 ?

Kudos to you! I couldn't figure out how to get the 20 so I had to do it the other way around.

Mindreko · **Posted:** Fri Nov 04, 2011 8:00 pm

You can work this by
Total Ways Of Selecting Commitee - ways of selecting NO women

Using Slot Method
10x9x8/3! - 6x5x4/3!

= 120 - 20

= 100

Hope that helps.

u0422811 · **Posted:** Sat Dec 03, 2011 11:41 am

Fairly simple.

First calculate the total of ways to choose 3 out of 10.

So this is 10!/3!(10-3)! which works out to 120.
Then you need to subtract the chances of getting no women - 6!/3!(6-3)! = 20.

Then 120-20 = 100.

Answer E.

Similar topics	Author	Replies	Views	Last post
combi 3 in GMAT Math Questions and Intellectual Discussions	stolyar	14	696	Thu Mar 06, 2003 12:47 am
combi 4 in GMAT Math Questions and Intellectual Discussions	stolyar	13	459	Thu Mar 06, 2003 12:58 am
simple combi in GMAT Math Questions and Intellectual Discussions	ashkg	11	389	Sun Jul 04, 2004 10:17 pm
Combi problem in GMAT Math Questions and Intellectual Discussions	ashkg	4	295	Mon Jun 13, 2005 7:09 am
PS - Combi in GMAT Problem Solving (PS)	karlfurt	6	243	Mon Nov 27, 2006 9:25 am

Go to GMAT Club Tests http://gmatclub.com/tests | What are GMAT Club Tests?

combi ques (m08q04)

Get Started:
Two Free Trial Tests

Get All GMATClub Tests

Who is online

Main Navigation

GMAT Resources

Partners

MBA Resources

Go to GMAT Club Tests http://gmatclub.com/tests | What are GMAT Club Tests?

combi ques (m08q04)

Get Started: Two Free Trial Tests

Get All GMATClub Tests

Who is online

Main Navigation

GMAT Resources

Partners

MBA Resources

Get Started:
Two Free Trial Tests