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combi ques (m08q04)

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combi ques (m08q04) [#permalink] New post 01 Aug 2008, 06:01
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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

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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 ?
[Reveal] Spoiler: OA
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Re: combi ques [#permalink] New post 01 Aug 2008, 06:05
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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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Re: combi ques [#permalink] New post 01 Aug 2008, 06:29
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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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Re: combi ques [#permalink] New post 01 Aug 2008, 08:32
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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Re: combi ques [#permalink] New post 01 Aug 2008, 08:46
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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Re: combi ques [#permalink] New post 01 Aug 2008, 09:04
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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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Re: combi ques [#permalink] New post 27 Oct 2010, 05:17
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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] New post 27 Oct 2010, 13:48
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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] New post 28 Oct 2010, 09:38
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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] New post 28 Oct 2010, 09:55
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] New post 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...

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Re: combi ques (m08q04) [#permalink] New post 04 Nov 2011, 19:00
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] New post 03 Dec 2011, 10:41
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] New post 05 Oct 2013, 07:15
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?

(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 ?


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

Thanks in advance..
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Re: combi ques (m08q04) [#permalink] New post 17 Oct 2013, 03:12
Expert's post
domfrancondumas wrote:
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?

(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 ?


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.

Does this make sense?

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Re: combi ques (m08q04) [#permalink] New post 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 ?
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Re: combi ques (m08q04) [#permalink] New post 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
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Re: combi ques (m08q04) [#permalink] New post 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.
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Re: combi ques (m08q04) [#permalink] New post 29 Oct 2013, 22:56
At least 1 means '1 or more'.

In this case, there are 4 women to choose from.

The possible methods of forming 'committees of 3' with AT LEAST 1 WOMAN are:

(1) 3 Women & 0 Men ———> 4C3 * 6C0 = 4 * 1 = 4

OR

(2) 2 Women & 1 Man ———> 4C2 * 6C1 = 6 * 6 = 36

OR

(3) 1 Woman & 2 Men ———> 4C1 * 6C2 = 4 * 15 = 60


So, there are 3 possible methods of forming committees. The 3 answers have to be added to arrive at the final answer.

Adding all the possible options --> 4 + 36 + 60 = 100

Answer is E
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Re: combi ques (m08q04) [#permalink] New post 02 Nov 2013, 12:42
thanks helped a lot
Re: combi ques (m08q04)   [#permalink] 02 Nov 2013, 12:42
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