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M08-04

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M08-04 [#permalink]

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New post 16 Sep 2014, 00:36
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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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Re M08-04 [#permalink]

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New post 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 \(C_3^{10} - C_3^6 = 120 - 20 = 100\).

Answer: E
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Re: M08-04 [#permalink]

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New post 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 \(C_3^{10} - C_3^6 = 120 - 20 = 100\).

Answer: E


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
\(C_4^1 * C_9^2 = 4 * 36 = 144\)

Thank you in advance for your answer,

Carnivor646
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Re: M08-04 [#permalink]

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New post 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 \(C_3^{10} - C_3^6 = 120 - 20 = 100\).

Answer: E


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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Re: M08-04 [#permalink]

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New post 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?
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Re: M08-04 [#permalink]

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New post 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
\(C_4^1 * C_9^2 = 4 * 36 = 144\)

Thank you in advance for your answer,

Carnivor646
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Re: M08-04 [#permalink]

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New post 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
\(C_4^1 * C_9^2 = 4 * 36 = 144\)

Thank you in advance for your answer,

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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M08-04 [#permalink]

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New post 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 :shock: :shock:

Thanks a tonne
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Re: M08-04 [#permalink]

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New post 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 :shock: :shock:

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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Re: M08-04 [#permalink]

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New post 08 Jan 2016, 09:21
oops!!
Thanks
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Re: M08-04 [#permalink]

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New post 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   [#permalink] 14 Nov 2017, 21:37
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