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A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
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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? (1) 635 (2) 700 (3) 1404 (4) 2620 (5) 3510
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Re: Commitee [#permalink]
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Geronimo 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?
(1) 635 (2) 700 (3) 1404 (4) 2620 (5) 3510 Method 2: 8 Men  5 Women Choose 6 ( two cases) Case 1:2 men, 4 women No of ways to choose 2 men out of 8 is 8C2 . This includes that one case in which the two men who refuse to serve together are included. So men can be chosen in 8C2  1 ways. No. of ways to choose 4 women out of 5 is 5C4. Number of ways of choosing 2 men and 4 women = (8C2  1)*(5C4) = 27*5 = 135 Case 2: 3 men, 3 women No of ways to choose 3 men out of 8 is 8C3. No of ways to choose those two men together is 6C1 (You choose them and choose one more to make 3). So men can be chosen in 8C3  6C1 ways. No of ways to choose 3 women out of 5 is 5C3. Number of ways of choosing 3 men and 3 women = (8C3  6C1)*(5C3) = 500 Total number of ways = 135 + 500 = 635 Case 2: 3 men, 3 women)
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Re: Commitee [#permalink]
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23 Jul 2012, 22:22
alphabeta1234 wrote:
Hey Bunuel,
Sorry I am in the same camp as Apex. I am trying two reconcile considering that I can pick 2 men, 3 women, and 1 of the remaining, instead of having to break it into two scenarios of (2 Men, 4 Women) and (3 Men, 4 Women)
Ways to chose 6 members committee without restriction: \(C^2_8*C^3_5*C^1_8 =2240\)
What I am doing wrong in the method above? The two should give the same number.
Be careful when you do not have to arrange the people but need to choose multiple times from the same group. Say, I want to make a team of 2 people from a group of 4 (A, B, C, D). In how many ways can I do it? Method1: 4C2 = 6 (AB, BC, CD, AC, AD, BD) Method 2: If instead, I select 1 and then another, this is how I will do it: 4C1 * 3C1 = 12 (AB, BA, AC, CA...) In the second case, I have arranged the 2 people in 2 ways. I first select A and then B in one case. I first select B and then A in another case. But a team doesn't need arrangement, it only needs selection. Hence, the second method is incorrect. Similarly, in this question, you need to make a committee i.e. only select, not arrange. When you pick 2 men, 3 women and 1 of the remaining, you are double counting. When you select 2 men in 8C2 ways, say you select A and B When you select 3 women in 5C3 ways, say you select X, Y and Z. When you select the last person, say you select a man C. Now consider another case: When you select 2 men in 8C2 ways, say you select A and C When you select 3 women in 5C3 ways, say you select X, Y and Z. When you select the last person, say you select a man B. The committee is same in both the cases (A, B, C, X, Y, Z) but we count them as 2 different cases when we use your method. Hence, there is double counting in your method. Therefore, you need to do what Bunuel suggested. For more on P&C, check my blog: http://www.veritasprep.com/blog/categor ... erwisdom/
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Re: Commitee [#permalink]
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17 Nov 2010, 08:23
prashantbacchewar wrote: VeritasPrepKarishma wrote: Method 2:
8 Men  5 Women Choose 6 ( two cases)
Case 1:2 men, 4 women No of ways to choose 2 men out of 8 is 8C2 . This includes that one case in which the two men who refuse to serve together are included. So men can be chosen in 8C2  1 ways. No. of ways to choose 4 women out of 5 is 5C4. Number of ways of choosing 2 men and 4 women = (8C2  1)*(5C4) = 27*5 = 135
Case 2: 3 men, 3 women No of ways to choose 3 men out of 8 is 8C3. No of ways to choose those two men together is 6C1 (You choose them and choose one more to make 3). So men can be chosen in 8C3  6C1 ways. No of ways to choose 3 women out of 5 is 5C3. Number of ways of choosing 3 men and 3 women = (8C3  6C1)*(5C3) = 500
Total number of ways = 135 + 500 = 635
Case 2: 3 men, 3 women)
Karishma I am not able to get the highlighted portion. Can you pls simplyfy for me. Sure Prashant. We are looking at the case where we take 3 men and 3 women. Now how do we choose 3 men? If there were no constraints, it would simply be 8C3. But, there are two men who do not want to be chosen together. Let us find out the opposite, i.e. in how many ways can we choose those two men together. Once we get this number, we can subtract it from 8C3 to get the number of ways of choosing 3 men out of 8 with the required constraints. Now, in how many ways can we choose those two men together. We take those two men and choose one more out of the remaining 6 using 6C1. Now we have chosen 3 men. So 8C3  6C1 gives the number of ways of choosing three men when those two men are not taken together.
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
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20 Feb 2012, 00:54
Apex231 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?
Number of ways to choose 2 men = 8C2 = 28 Number of ways to choose 3 women = 5C3 = 10 Total ways to choose 2 men and 3 women = 28 * 10 = 280
Now the remaining committee member can be chosen from the 8 remaining men and women. Total ways to choose 6 committee members = 280 * 8 = 2240
When the two men who refuse to serve together are selected  Number of ways to choose 3 women = 5C3 = 10 Remaining 1 committee member can be chosen from the 8 remaining men and women. Total combinations = 10 * 8 = 80
Total different committees if two of the men refuse to serve together = 2240  80 = 2160
I am doing something wrong here, where am i going wrong?

Realize my mistake, i am duplicating count when i multiply by 8. The numbers you get will have duplications. Let's take for example the red part of your solution above: consider five women: {A, B, C, D, E}. When you choose 3 of them (with \(C^3_5\)) you can get for example the group {A, B, C} next when you choose one from 8 people then you can get one more woman, for example D, so you'll have in the group 4 women {A, B, C, D}. Now, if you choose the group {A, B, D}, with \(C^3_5\) and then choose C from 8 people then you'll basically will get the same 4 women group: {A, B, C, D}. Hope it's clear.
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Re: Commitee [#permalink]
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17 Nov 2010, 07:16
VeritasPrepKarishma wrote: Method 2:
8 Men  5 Women Choose 6 ( two cases)
Case 1:2 men, 4 women No of ways to choose 2 men out of 8 is 8C2 . This includes that one case in which the two men who refuse to serve together are included. So men can be chosen in 8C2  1 ways. No. of ways to choose 4 women out of 5 is 5C4. Number of ways of choosing 2 men and 4 women = (8C2  1)*(5C4) = 27*5 = 135
Case 2: 3 men, 3 women No of ways to choose 3 men out of 8 is 8C3. No of ways to choose those two men together is 6C1 (You choose them and choose one more to make 3). So men can be chosen in 8C3  6C1 ways. No of ways to choose 3 women out of 5 is 5C3. Number of ways of choosing 3 men and 3 women = (8C3  6C1)*(5C3) = 500
Total number of ways = 135 + 500 = 635
Case 2: 3 men, 3 women)
Karishma I am not able to get the highlighted portion. Can you pls simplyfy for me.



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Re: Commitee [#permalink]
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18 Nov 2010, 07:52
This is what I still need to improve... Thx for both answers !!



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Re: A committee of 6 [#permalink]
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19 Feb 2012, 07:30
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Well, i do not get the 635 but narrowed it down to answer E
Total (favourable) commitees = All possible commitees  restriction
Option 1: M M W W W W (2 Men & 4 Women) Option 2: M M M W W W (3 Men & 3 Women) Restriction: 2 out of 8 Men refuse to serve together
Outcomes Option 1: \(\frac{8!}{2!*6!} * \frac{5!}{4!*1!}\) = 28 * 5 = 140
Outcomes Option 2: \(\frac{8!}{3!*5!} * \frac{5!}{3!*2!}\) = 56 * 10 = 560
Total commitees = 700
Outcomes restriction option 1: 1 (if 2 men refuse to serve together and 2 places are available) * 5 (Outcomes women) = 5 Outcomes restriction option 2: 3 (if 2 men refuse to serve together and 3 places are available) * 10 (Outcomes women = 30
So it should be definitely less than 700 but i guess i made a mistake somewhere. POE E



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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
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19 Feb 2012, 08:50
Thanks Bunuel, i have found my mistake. I did not consider that the space left (at the restriction part) can be filled with all 6 remaining "men". I did just consider the possibilities to fill the places. Silly!!



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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
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19 Feb 2012, 09:00
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?
Number of ways to choose 2 men = 8C2 = 28 Number of ways to choose 3 women = 5C3 = 10 Total ways to choose 2 men and 3 women = 28 * 10 = 280
Now the remaining committee member can be chosen from the 8 remaining men and women. Total ways to choose 6 committee members = 280 * 8 = 2240
When the two men who refuse to serve together are selected  Number of ways to choose 3 women = 5C3 = 10 Remaining 1 committee member can be chosen from the 8 remaining men and women. Total combinations = 10 * 8 = 80
Total different committees if two of the men refuse to serve together = 2240  80 = 2160
I am doing something wrong here, where am i going wrong?

Realize my mistake, i am duplicating count when i multiply by 8.



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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
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20 Feb 2012, 08:36
Thanks Bunuel for explaining..it's clear to me now..



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Re: Commitee [#permalink]
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22 Jul 2012, 20:19
Bunuel wrote: Geronimo 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?
(1) 635 (2) 700 (3) 1404 (4) 2620 (5) 3510 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: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\) 70065 = 635 Answer: A. Hey Bunuel, Sorry I am in the same camp as Apex. I am trying two reconcile considering that I can pick 2 men, 3 women, and 1 of the remaining, instead of having to break it into two scenarios of (2 Men, 4 Women) and (3 Men, 4 Women) Ways to chose 6 members committee without restriction: \(C^2_8*C^3_5*C^1_8 =2240\) What I am doing wrong in the method above? The two should give the same number.



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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
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25 Jul 2012, 04:23
Elegant solutions provided above; however, please let me know where did I go wrong.
8 Men, can be considered as 6men and 2men, further we have 5 women to choose from:
Thus, at least 2 and 3 mean;
Case 1: 2 men and 4 women: 6C1*2C1*5C4 + 6C2*5C4
Case 2: 3 Men and 3 Women: 6C2*2C1*5C3 + 6C3*5C3
I don't the same number of ways. What am I doing wrong?



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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
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25 Jul 2012, 06:15
pavanpuneet wrote: Elegant solutions provided above; however, please let me know where did I go wrong.
8 Men, can be considered as 6men and 2men, further we have 5 women to choose from:
Thus, at least 2 and 3 mean;
Case 1: 2 men and 4 women: 6C1*2C1*5C4 + 6C2*5C4
Case 2: 3 Men and 3 Women: 6C2*2C1*5C3 + 6C3*5C3
I don't the same number of ways. What am I doing wrong? You are absolutely right. Why do you think you are wrong? Check the expressions, they will give you the correct answer.
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Re: Commitee [#permalink]
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23 May 2013, 20:28
Bunuel wrote: Geronimo 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?
(1) 635 (2) 700 (3) 1404 (4) 2620 (5) 3510 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: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\) 70065 = 635 Answer: A. Been trying to figure this one out, but I'm not getting your calculations when choosing it with the two men together. How do you arrive at \(C^2_2\) and the \((C^2_2*C^1_6)\) . Sorry if I'm being a bit slow on this one, but just haven't wrapped my brain around this.



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Re: Commitee [#permalink]
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24 May 2013, 01:39
Dixon wrote: Bunuel wrote: Geronimo 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?
(1) 635 (2) 700 (3) 1404 (4) 2620 (5) 3510 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: \(C^2_2*C^4_5+(C^2_2*C^1_6)*C^3_5=5+60=65\) 70065 = 635 Answer: A. Been trying to figure this one out, but I'm not getting your calculations when choosing it with the two men together. How do you arrive at \(C^2_2\) and the \((C^2_2*C^1_6)\) . Sorry if I'm being a bit slow on this one, but just haven't wrapped my brain around this. 2 men and 4 women: \(C^2_2*C^4_5=1*5\). One way to choose 2 particular men out of 2: \(C^2_2=1\). 3 men and 3 women: \((C^2_2*C^1_6)*C^3_5\). One way to choose 2 particular men out of 2: \(C^2_2=1\) and 6 ways to choose the third men out of the remaining 6: \(C^1_6\). Hope it's clear.
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
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Bunuel wrote: 2 men and 4 women: \(C^2_2*C^4_5=1*5\). One way to choose 2 particular men out of 2: \(C^2_2=1\).
3 men and 3 women: \((C^2_2*C^1_6)*C^3_5\). One way to choose 2 particular men out of 2: \(C^2_2=1\) and 6 ways to choose the third men out of the remaining 6: \(C^1_6\).
Hope it's clear.
Hi Bunuel, Sorry but I am still not clear why you have split the selection of men into 2,1 form!! \((C^2_2*C^1_6)*C^3_5\). One way to choose 2 particular men out of 2: \(C^2_2=1\) and 6 ways to choose the third men out of the remaining 6: \(C^1_6\)
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Re: A committee of 6 is chosen from 8 men and 5 women so as to [#permalink]
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22 Aug 2014, 04:48
scofield1521 wrote: Bunuel wrote: 2 men and 4 women: \(C^2_2*C^4_5=1*5\). One way to choose 2 particular men out of 2: \(C^2_2=1\).
3 men and 3 women: \((C^2_2*C^1_6)*C^3_5\). One way to choose 2 particular men out of 2: \(C^2_2=1\) and 6 ways to choose the third men out of the remaining 6: \(C^1_6\).
Hope it's clear.
Hi Bunuel, Sorry but I am still not clear why you have split the selection of men into 2,1 form!! \((C^2_2*C^1_6)*C^3_5\). One way to choose 2 particular men out of 2: \(C^2_2=1\) and 6 ways to choose the third men out of the remaining 6: \(C^1_6\) We are counting the number of committees with 3 men and 3 women, where out of these 3 men 2 are fixed, so basically we are only selecting 1 man out of the remaining 6 men. 2C2 is the number of ways to select those two fixed men out of 2, so that's 1 way to choose and 6C1 is the number of ways to select 1 men out of 6.
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