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# M08-04

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Math Expert
Joined: 02 Sep 2009
Posts: 49987

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16 Sep 2014, 00:36
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Difficulty:

25% (medium)

Question Stats:

71% (01:29) correct 29% (00:56) wrong based on 110 sessions

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If 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

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Math Expert
Joined: 02 Sep 2009
Posts: 49987

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16 Sep 2014, 00:36
2
Official Solution:

If 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

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is $$C_3^{10}$$. From this figure we have to subtract the number of committees that consist entirely of men: $$C_3^6$$. The final answer is $$C_3^{10} - C_3^6 = 120 - 20 = 100$$.

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Joined: 07 Aug 2015
Posts: 5

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17 Aug 2015, 02:02
Bunuel wrote:
Official Solution:

If 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

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is $$C_3^{10}$$. From this figure we have to subtract the number of committees that consist entirely of men: $$C_3^6$$. The final answer is $$C_3^{10} - C_3^6 = 120 - 20 = 100$$.

Hello Bunuel,

I think that there is an error with the way you edited the combinaisions which are inverted meaning that you're picking 6 men out 3 ($$C_3^6$$) which is impossible (am I right?).

Also I would have liked to know if there was another way of solving this problem by "fixing" one women and then choosing two other members out of the nine remaining. I tried to do so using this formula but I didn't get the right solution:
# of ways to select one women out of four = 4
# of ways to select two other members out of nine = 36
$$C_4^1 * C_9^2 = 4 * 36 = 144$$

Carnivor646
Math Expert
Joined: 02 Sep 2009
Posts: 49987

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17 Aug 2015, 04:20
Carnivor646 wrote:
Bunuel wrote:
Official Solution:

If 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

Consider an unconstrained version of the question: in how many ways can a committee of 3 be formed? The answer is $$C_3^{10}$$. From this figure we have to subtract the number of committees that consist entirely of men: $$C_3^6$$. The final answer is $$C_3^{10} - C_3^6 = 120 - 20 = 100$$.

Hello Bunuel,

I think that there is an error with the way you edited the combinaisions which are inverted meaning that you're picking 6 men out 3 ($$C_3^6$$) which is impossible (am I right?).

$$C^3_6$$, $$C^6_3$$, 6C3 are the same: choosing 3 out of 6. How can we choose 6 out of 3? Those are just different ways of writing the same.
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Joined: 10 Aug 2014
Posts: 49
GMAT 1: 680 Q49 V34

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06 Sep 2015, 14:26
Bunuel,
i tried to solve problem by using this formula:
1C4 - choosing 1 woman
2C4 - 2 woman out 4
3C4 - 3 woman out of 4

2C6 - 2 man out of 6
1C6 - 1 man out of 6

1C4+2C6+2C4+1C6+3C4=4+15+6+6+4=35
Can you please show me where i went wrong?
Intern
Joined: 07 Aug 2015
Posts: 5

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07 Sep 2015, 02:11
Hello Bunuel,

Could you tell why my solution of "fixing" one women and then choosing two other members out of the nine remaining didn't work please? As I said, I tried to do so using this formula but my solution did not match yours:

# of ways to select one women out of four = 4
# of ways to select two other members out of nine = 36
$$C_4^1 * C_9^2 = 4 * 36 = 144$$

Carnivor646
Math Expert
Joined: 02 Sep 2009
Posts: 49987

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07 Sep 2015, 03:19
1
Carnivor646 wrote:
Hello Bunuel,

Could you tell why my solution of "fixing" one women and then choosing two other members out of the nine remaining didn't work please? As I said, I tried to do so using this formula but my solution did not match yours:

# of ways to select one women out of four = 4
# of ways to select two other members out of nine = 36
$$C_4^1 * C_9^2 = 4 * 36 = 144$$

Carnivor646

The number you get will have duplications.

The number you get will have duplication.

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.
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Joined: 03 Jul 2015
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08 Jan 2016, 07:55
I have another solution. Clearly something is wrong with it.
Bunuel, Could you Please tell me why ? For my understanding only .

So if we have to select 3 people from 4 women 6 men ; at least one of them has to be a woman.
We can select in 3 ways :
1 woman 2 men : 4C1 X 6C2
2 women 1 man : 4C2 X 6C1
3 women : 4C3
We ad them all : 19+36+4 = 59

Thanks a tonne
Intern
Joined: 22 Apr 2015
Posts: 12
GMAT 1: 760 Q50 V44

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08 Jan 2016, 09:14
shreyashid wrote:
I have another solution. Clearly something is wrong with it.
Bunuel, Could you Please tell me why ? For my understanding only .

So if we have to select 3 people from 4 women 6 men ; at least one of them has to be a woman.
We can select in 3 ways :
1 woman 2 men : 4C1 X 6C2
2 women 1 man : 4C2 X 6C1
3 women : 4C3
We ad them all : 19+36+4 = 59

Thanks a tonne

shreyashid

You're actually right, but your math is wrong.
4C1 = 4 and 6C2 = 15

So you should be adding up 60+36+4 = 100.

However, you should also note that the published solution takes the least amount of calculation (2 choose functions vs 6), meaning that there is much more room for error!

Cheers
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Joined: 03 Jul 2015
Posts: 29

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08 Jan 2016, 09:21
oops!!
Thanks
Intern
Joined: 25 Jul 2017
Posts: 2
Location: India
Schools: ISB '20
GMAT 1: 710 Q49 V39
GPA: 3.75

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14 Nov 2017, 21:37
mika84 wrote:
Bunuel,
i tried to solve problem by using this formula:
1C4 - choosing 1 woman
2C4 - 2 woman out 4
3C4 - 3 woman out of 4

2C6 - 2 man out of 6
1C6 - 1 man out of 6

1C4+2C6+2C4+1C6+3C4=4+15+6+6+4=35
Can you please show me where i went wrong?

The approach is correct but the way you applied it isn't. Note the distinction between AND and OR. Let's say we have three spaces to fill.

A B C

Case 1:

Two men "AND" one woman. Whenever we have "AND", we multiply.

So, 6C2 * 4 = 60

Case 2:

One man and two women.

6 * 4C2 = 36

Case 3:

All 3 women.

4C3 = 4

Now, we can have either of these choices. MMF "OR" MFF "OR" FFF. We add them in case of "OR". That gives us:

60+36+4 = 100

Simply put, multiply when you have AND and add when you have OR. Hope this helps.
Re: M08-04 &nbs [#permalink] 14 Nov 2017, 21:37
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# M08-04

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