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A committee of 6 is chosen from 8 men and 5 women so as to

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A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 29 Jul 2010, 19:36
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A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635
[Reveal] Spoiler: OA
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Re: Another Combination Problem [#permalink] New post 29 Jul 2010, 19:41
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jakolik wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635


Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): C^2_8*C^4_5+C^3_8*C^3_5 = 700

Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65

700-65 = 635

Answer: E.
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Re: Another Combination Problem [#permalink] New post 29 Jul 2010, 20:33
Thank you Bunuel.

Can you please explain why ways to chose 6 members committee without restriction cannot be: C^2_8*C^3_5*C^1_8
We are to choose 2 men out of 8, 3 women out of 5 and 1 (woman or man) out of 8 (13-5)
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Re: Another Combination Problem [#permalink] New post 29 Jul 2010, 21:05
jakolik wrote:
Thank you Bunuel.

Can you please explain why ways to chose 6 members committee without restriction cannot be: C^2_8*C^3_5*C^1_8
We are to choose 2 men out of 8, 3 women out of 5 and 1 (woman or man) out of 8 (13-5)


This way there will be duplicates in the combinations.

Consider M1,M2....M8 be men and W1,W2......W5 be women
See the below 3 combinations. They are in fact same but will be counted as different with the approach you mentioned

...2M.. .....3W...... 1M or 1W
M1,M2| W1,W2,W3| M3
M3,M2| W1,W2,W3| M1
M3,M1| W1,W2,W3| M2

Thanks
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Re: Another Combination Problem [#permalink] New post 31 Jul 2010, 08:41
Quote:
Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65


Bunuel, can you explain the logic behind this?
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Re: Another Combination Problem [#permalink] New post 15 Sep 2012, 06:04
Yups Like others I too have the same query

Anybody explain the logic selecting two men who work together with each other , please?

We only have the symbols here but nothing explaining the same ?
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Re: Another Combination Problem [#permalink] New post 15 Sep 2012, 07:28
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stne wrote:
Yups Like others I too have the same query

Anybody explain the logic selecting two men who work together with each other , please?

We only have the symbols here but nothing explaining the same ?


Quote:
Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65


Denote by A and B the two particular men.
We can have 2 men and 4 women chosen, when the two men are the particular two: choose the two men C^2_2, then choose 4 women out of the 5, which is C^4_5. So, A,B and four women chosen.
We can also have the two men chosen, again C^2_2, choose another man from the remaining 6 men, this is C^1_6, then choose 3 women out of the 5, which is C^3_5. Here, A, B, another men, and 3 women are chosen.
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 15 Sep 2012, 07:48
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jakolik wrote:
Thank you Bunuel.

Can you please explain why ways to chose 6 members committee without restriction cannot be: C^2_8*C^3_5*C^1_8
We are to choose 2 men out of 8, 3 women out of 5 and 1 (woman or man) out of 8 (13-5)


Bunuel wrote:
jakolik wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635


Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): C^2_8*C^4_5+C^3_8*C^3_5 = 700

Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65

700-65 = 635

Answer: E.


Committee has following formations-
M----------W
2----------4
3----------3

Quote:
Ways to chose 6 members committee without restriction (two men refuse to server together): C^2_8*C^4_5+C^3_8*C^3_5 = 700


This value 700 includes committees in which both of the men (say X and Y) are included. So, we need to find out number of such committees and deduct those from the total 700.

In the first case, 2 men are chosen. No. of committees of 2 men in which X and Y are included is 1
So, Considering women also we find total such committees= 1 * ^5C_4 = 5

In the second case, 3 men are chosen out of 6.
Also, out of 3 men X and Y will always be in the committee. Therefore, we need to find out 1 more man out of remaining 6. This can be done ^6C_1way = 6 ways

So, Considering women also we find total such committees= 6 * ^5C_3 = 6*10

Therefore, total no. of committees in which both X and Y are included is 5+60=65

This gives us answer, total no. of committees in which both X and Y are not included= 700-65=635
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 24 Sep 2012, 03:10
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Find out how many different combinations can be made between the 8 men and 5 women ...

We will have two scenarios , First where we have 2 men and 4 women , and second where we have 3 men and 4 women ... Finding the combinations for each and then adding them up gives us 700 ...

Logically the number of combinations where any 2 men do not serve together would have to be lower than 700 , therefore the only option is E ....
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Re: Another Combination Problem [#permalink] New post 17 Oct 2013, 09:19
Bunuel wrote:
jakolik wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635


Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): C^2_8*C^4_5+C^3_8*C^3_5 = 700

Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65

700-65 = 635

Answer: E.


Hi,
I went at it a different way and did not get the right result.
Can you find out where the logic is wrong?

Total number of possibilities: choose 2 men, choose 3 women, choose one from the leftover: 8c2*5c3*8c1
Total number of possibilities for the 2 men together: 2C2*5C3*8C1.
And last step, subtract the two numbers.... This is not working for me... Why?
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 17 Oct 2013, 09:44
jakolik wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635


Bunuel's subtraction way works. Here's another method -

There are 2 men and 3 women leaving out one spot.

Case 1: We have 3 men and 3 women. Total number of ways = (8C3 - 6)* 5C3 = 500
Case 2: We have 2 men and 4 women. Total number of ways = (8C2 - 1)*5C4 = 135

Total = 635
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink] New post 18 Oct 2013, 02:06
jeopardo wrote:
jakolik wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635


Bunuel's subtraction way works. Here's another method -

There are 2 men and 3 women leaving out one spot.

Case 1: We have 3 men and 3 women. Total number of ways = (8C3 - 6)* 5C3 = 500
Case 2: We have 2 men and 4 women. Total number of ways = (8C2 - 1)*5C4 = 135

Total = 635

I understand that Bunuels way works.... My question is why my way doesn't?
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Re: Another Combination Problem [#permalink] New post 20 Oct 2013, 12:23
Expert's post
ronr34 wrote:
Bunuel wrote:
jakolik wrote:
A committee of 6 is chosen from 8 men and 5 women so as to contain at least 2 men and 3 women. How many different committees could be formed if two of the men refuse to serve together?

(A) 3510
(B) 2620
(C) 1404
(D) 700
(E) 635


Committee can have either: 2 men and 4 women OR 3 men and 3 women (to meet the condition of at least 2 men and 3 women).

Ways to chose 6 members committee without restriction (two men refuse to server together): C^2_8*C^4_5+C^3_8*C^3_5 = 700

Ways to chose 6 members committee with two particular men serve together: C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65

700-65 = 635

Answer: E.


Hi,
I went at it a different way and did not get the right result.
Can you find out where the logic is wrong?

Total number of possibilities: choose 2 men, choose 3 women, choose one from the leftover: 8c2*5c3*8c1
Total number of possibilities for the 2 men together: 2C2*5C3*8C1.
And last step, subtract the two numbers.... This is not working for me... Why?


Check here: 4-professors-and-6-students-are-being-considered-for-85865-20.html#p1265096
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RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

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: Another Combination Problem   [#permalink] 20 Oct 2013, 12:23
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