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Combinatory questions

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Combinatory questions

by MarcoRome » Sun Aug 22, 2010 3:35 am
Among 16 people, 7 are male and 9 are female. If each group is formed by three people and each group has to contain at least 1 male and 1 female, how many groups can be formed?

I solved it in the following way, but I am not sure it is right:
(Total number of 3-people groups) - (3-male groups) - (3-female groups)
total number of 3-people groups: (16 3)= 16!/(13!*3!)= 16*15*14/3*2=8*5*14=560
3-male groups: (7 3)= 7!/(4!*3!)= 7*6*5/3*2=35
3-female groups: (9 3)= 9!/(6!*3!)=9*8*7/3*2=3*4*7=84
560-35-84=441

is it correct?

Marco
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by sirisha.g » Sun Aug 22, 2010 4:41 am
MarcoRome wrote:Among 16 people, 7 are male and 9 are female. If each group is formed by three people and each group has to contain at least 1 male and 1 female, how many groups can be formed?

I solved it in the following way, but I am not sure it is right:
(Total number of 3-people groups) - (3-male groups) - (3-female groups)
total number of 3-people groups: (16 3)= 16!/(13!*3!)= 16*15*14/3*2=8*5*14=560
3-male groups: (7 3)= 7!/(4!*3!)= 7*6*5/3*2=35
3-female groups: (9 3)= 9!/(6!*3!)=9*8*7/3*2=3*4*7=84
560-35-84=441

is it correct?

Marco
Hello Marco,

This approach is not right way of doing a P & C problem. This approach is used in probability where P(at least one)=1-P(one).

For solving this problem you need to follow the below method.

atleast 1 male and 1 female => 1 male and 1 female should definitely be selected => there can be 2 males and 1 female or 2 females and 1 male

therefore case1: you can select 1 male in 7C1 ways. 2nd male in 6C1 ways and 1 female in 9C1 ways =7*6*9=378
case2: you can select 1 female in 9C1 ways, 2nd in 8C1 ways and 1 male in 7C1 ways=9*8*7=504

either case 1 or case 2= 378+504=882
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by HPengineer » Sun Aug 22, 2010 6:27 pm
Shouldnt the combinations by dividied by two factorial? since Male Male Female is the same is Female Male Male??

Men = 7*6*9/2!
Women 9*8*7/2!
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by sirisha.g » Sun Aug 22, 2010 7:14 pm
HPengineer wrote:Shouldnt the combinations by dividied by two factorial? since Male Male Female is the same is Female Male Male??

Men = 7*6*9/2!
Women 9*8*7/2!
Order doesn't matter because persons are different
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by limestone » Sun Aug 22, 2010 7:29 pm
In think they are asking about how many groups not how many ways that group can be formed. So we can solve this by either eliminating the two disqualified cases (all male or all female) or counting only accepting cases ( 2 males + 1 female, or 2 female + 1 male)

First method:
Total group: 16C3 = 560
3 Males group: 7C3 = 35
3 Females group: 9C3 = 84
No. of Qualified group: 560-35-84 = 441

Method 2:
2 M + 1 F : 7C2 * 9C1 = 21 * 9 = 189
1 M + 2 F : 7C1 * 9C2 = 7 * 36 = 252
No. of Qualified group: 189 + 252 = 441

Same result
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by rohit_gmat » Sun Aug 22, 2010 11:20 pm
sirisha.g wrote: therefore case1: you can select 1 male in 7C1 ways. 2nd male in 6C1 ways and 1 female in 9C1 ways =7*6*9=378
case2: you can select 1 female in 9C1 ways, 2nd in 8C1 ways and 1 male in 7C1 ways=9*8*7=504

either case 1 or case 2= 378+504=882
Why is the above wrong?

To simplify it - lets say there are 10 Males and we need to make a group of 2 Males and the question is in how many ways can we do so?
The normal approach would be 10C2 = 10!/8!2! = 10 x 9 /2 = 45

But what if we look at it in this way =>
Choices for the 1st Male would be 10C1 = 10!/9! = 10
Choices for the 2nd Male would be 9C1 = 9!/8! = 9
Total would be 10 x 9 = 90

double of 45!

Whats wrong in the 2nd logic and why do we have different answers?

Please help.

Thanks
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by Stuart@KaplanGMAT » Sun Aug 22, 2010 11:24 pm
rohit_gmat wrote:
sirisha.g wrote: therefore case1: you can select 1 male in 7C1 ways. 2nd male in 6C1 ways and 1 female in 9C1 ways =7*6*9=378
case2: you can select 1 female in 9C1 ways, 2nd in 8C1 ways and 1 male in 7C1 ways=9*8*7=504

either case 1 or case 2= 378+504=882
Why is the above wrong?

To simplify it - lets say there are 10 Males and we need to make a group of 2 Males and the question is in how many ways can we do so?
The normal approach would be 10C2 = 10!/8!2! = 10 x 9 /2 = 45

But what if we look at it in this way =>
Choices for the 1st Male would be 10C1 = 10!/9! = 10
Choices for the 2nd Male would be 9C1 = 9!/8! = 9
Total would be 10 x 9 = 90

double of 45!

Whats wrong in the 2nd logic and why do we have different answers?

Please help.

Thanks
The problem with the second approach is that you're pretending that order matters, when it doesn't.

Let's call our 10 people A, B, C, D, E, F, G, H, I and J.

Using your approach, one pair could be A then B; another pair could be B then A. However, since we just care about who ends up in the group, AB is the exact same pairing as BA - you've counted them twice.
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by Stuart@KaplanGMAT » Sun Aug 22, 2010 11:32 pm
MarcoRome wrote:Among 16 people, 7 are male and 9 are female. If each group is formed by three people and each group has to contain at least 1 male and 1 female, how many groups can be formed?

I solved it in the following way, but I am not sure it is right:
(Total number of 3-people groups) - (3-male groups) - (3-female groups)
total number of 3-people groups: (16 3)= 16!/(13!*3!)= 16*15*14/3*2=8*5*14=560
3-male groups: (7 3)= 7!/(4!*3!)= 7*6*5/3*2=35
3-female groups: (9 3)= 9!/(6!*3!)=9*8*7/3*2=3*4*7=84
560-35-84=441

is it correct?

Marco
One problem is that the question is horribly worded - you'd never see such ambiguous language in an actual GMAT question - what's the source?

Let's clean it up to better express (what I think is) the intended meaning:
There are 7 men and 9 women available to form one group of 3 people. If the group must contain at least one man and at least one woman, how many different groups can be created?
As noted by sirisha, the best way to solve is to break the problem down into two cases (although sirisha's computations weren't correct, the idea was good):

1) 2 men and 1 woman; and
2) 1 man and 2 women.

Since we want case (1) OR case (2), we then add the results together.

1) 7C2 * 9C1 = (7*6/2) * 9 = 21 * 9 = 189

2) 7C1 * 9C2 = 7 * (9*8/2) = 7 * 36 = 252

Total number of possible groups = 189 + 252 = 441

Marco, your approach also works, but in this particular case it's more work to subtract what you don't want than to just add up what you do want.
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by rohit_gmat » Sun Aug 22, 2010 11:39 pm
rohit_gmat wrote:
sirisha.g wrote: therefore case1: you can select 1 male in 7C1 ways. 2nd male in 6C1 ways and 1 female in 9C1 ways =7*6*9=378
case2: you can select 1 female in 9C1 ways, 2nd in 8C1 ways and 1 male in 7C1 ways=9*8*7=504

either case 1 or case 2= 378+504=882
Why is the above wrong?

To simplify it - lets say there are 10 Males and we need to make a group of 2 Males and the question is in how many ways can we do so?
The normal approach would be 10C2 = 10!/8!2! = 10 x 9 /2 = 45

But what if we look at it in this way =>
Choices for the 1st Male would be 10C1 = 10!/9! = 10
Choices for the 2nd Male would be 9C1 = 9!/8! = 9
Total would be 10 x 9 = 90

double of 45!

Whats wrong in the 2nd logic and why do we have different answers?

Please help.

Thanks
On a second thought... i think the second approach is wrong because it considers sequence (is permutation rather than combination), so we need to divide it by 2! (2)....?
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by rohit_gmat » Mon Aug 23, 2010 12:02 am
Ahh! Now I get it !!!
Just got your updated posts right now...

Thanks a lot Stuart !
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by MarcoRome » Mon Aug 23, 2010 1:33 am
guys, thanks a lot! the rigth solution is 441. and the puzzle here was to distinguish between permutation and combination.

if you have other questions similar to these ones, please let me know. Combinations/Permutations are my weakeness.

Ciao!
Marco
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