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combi ques (m08q04) [#permalink]
 01 Aug 2008, 07:01
Question Stats:
77% (01:42) correct
22% (01:35) wrong based on 57 sessions
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 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 ?
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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 ?
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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
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thanks guys..
But what exactly am I doing wrong with my method ?
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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
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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
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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
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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
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Re: combi ques (m08q04) [#permalink]
27 Oct 2010, 07:37
Easy E...10c3 - 6c3
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Re: combi ques (m08q04) [#permalink]
27 Oct 2010, 14:48
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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.
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Re: combi ques (m08q04) [#permalink]
28 Oct 2010, 10:38
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?
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Re: combi ques (m08q04) [#permalink]
28 Oct 2010, 10:55
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.
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Re: combi ques (m08q04) [#permalink]
29 Oct 2010, 03:12
10C3-6C3..this is simplest way of doing the problem...
Answer : E
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Re: combi ques (m08q04) [#permalink]
31 Oct 2011, 07:02
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...
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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 
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Re: combi ques (m08q04) [#permalink]
01 Nov 2011, 00:23
This is very easy 10c3 -6c3 = 120-20 =100
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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.
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Re: combi ques (m08q04) [#permalink]
04 Nov 2011, 20:00
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.
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Re: combi ques (m08q04) [#permalink]
03 Dec 2011, 11:41
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.
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Re: combi ques (m08q04)
[#permalink]
03 Dec 2011, 11:41
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