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# a club has 8 male and 8 female members. the club is choosing

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VP
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a club has 8 male and 8 female members. the club is choosing [#permalink]  02 Mar 2005, 14:21
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a club has 8 male and 8 female members. the club is choosing a commitee of 6 members. the commitee must have 3 male and 3 female members. how many different commitees must be chosen ?
SVP
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[#permalink]  02 Mar 2005, 14:30
Want to make it a little harder?;)

1. A club has 8 male and 8 female members. The club is choosing a commitee of 6 members. The commitee must have at least 2 male and 2 female members. How many different commitees must be chosen ?

2. A club has 8 male including Robert and 8 female members including Ann. The club is choosing a commitee of 6 members. The commitee must have 3 male and 3 female members. Robert and Ann refused to serve in the same committee. How many different commitees must be chosen ?
VP
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[#permalink]  02 Mar 2005, 14:36
HongHu wrote:
Want to make it a little harder?;)

1. A club has 8 male and 8 female members. The club is choosing a commitee of 6 members. The commitee must have at least 2 male and 2 female members. How many different commitees must be chosen ?

2. A club has 8 male including Robert and 8 female members including Ann. The club is choosing a commitee of 6 members. The commitee must have 3 male and 3 female members. Robert and Ann refused to serve in the same committee. How many different commitees must be chosen ?

1. 8C2*8C2*12C2

2. 8C3*8C3 - 1*7C2*7C2......(total ways - ways where they r always together)
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[#permalink]  02 Mar 2005, 14:52
Totally agree with banerjeea_98 with all answers. I liked his approach to question #1 of Honghu's question. I calculated the ind. possibilities with 2M+4F, 3M+3F and 4M+2F and it's longer. Good approach baneerja
_________________

Best Regards,

Paul

VP
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[#permalink]  02 Mar 2005, 15:03
hmm ? what is wrong with these answers ?

1. 2 men and 4 women = 8c2*8c4 and 3 men and 3 women = 8c3 * 8c3 and 4 men and 2 women = 8c4 * 8c2 => add them

2. 7c3 (without robert) * 8c3 (with ann) + 7c3 (without ann) * 8c3 (with robert)
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[#permalink]  02 Mar 2005, 15:39
Good job baner!

christoph wrote:
hmm ? what is wrong with these answers ?

1. 2 men and 4 women = 8c2*8c4 and 3 men and 3 women = 8c3 * 8c3 and 4 men and 2 women = 8c4 * 8c2 => add them

2. 7c3 (without robert) * 8c3 (with ann) + 7c3 (without ann) * 8c3 (with robert)

I believe your solution for 1 is correct. But you've got to picked up a lot repetitions for your second solution. The first 7C3*8C3 have included outcomes that have Ann and don't have Ann. And the second 7C3*8C3 have included outcomes that have Robert and don't have Robert. Does this make sense to you? Perhaps you could calculate the numbers out to compare the answers from the different approaches. When I have time in my test, I'd do two approaches to verify my answer too.
Manager
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[#permalink]  02 Mar 2005, 16:54
banerjeea_98 wrote:
HongHu wrote:
Want to make it a little harder?;)

1. A club has 8 male and 8 female members. The club is choosing a commitee of 6 members. The commitee must have at least 2 male and 2 female members. How many different commitees must be chosen ?

2. A club has 8 male including Robert and 8 female members including Ann. The club is choosing a commitee of 6 members. The commitee must have 3 male and 3 female members. Robert and Ann refused to serve in the same committee. How many different commitees must be chosen ?

1. 8C2*8C2*12C2

2. 8C3*8C3 - 1*7C2*7C2......(total ways - ways where they r always together)

Why did you do 1*7C2*7C2? Was there something that like 8C8 that turned into 1?
VP
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[#permalink]  02 Mar 2005, 17:12
there is only 1 way to choose both Rob and Ann and then 7C2*7C2 ways to choose for the rest of the 4 positions.
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[#permalink]  02 Mar 2005, 17:20
banerjeea_98 wrote:
HongHu wrote:
Want to make it a little harder?;)

1. A club has 8 male and 8 female members. The club is choosing a commitee of 6 members. The commitee must have at least 2 male and 2 female members. How many different commitees must be chosen ?

2. A club has 8 male including Robert and 8 female members including Ann. The club is choosing a commitee of 6 members. The commitee must have 3 male and 3 female members. Robert and Ann refused to serve in the same committee. How many different commitees must be chosen ?

1. 8C2*8C2*12C2

2. 8C3*8C3 - 1*7C2*7C2......(total ways - ways where they r always together)

Hi Banerjee

Could you plz xplain me how you got 1) 8C2*8C2*12C2

Thanks,
Vijo
VP
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[#permalink]  02 Mar 2005, 17:23
sure....u have to choose atleast 2 men and 2 women...so let's do that first....8C2*8C2......now that we met that condition.....we r now left with 2 more positions in the committee.....and we can choose these 2 ppl from 12 remaining ppl.....12C2.....so total ways....8C2*8C2*12C2
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[#permalink]  02 Mar 2005, 17:26
Oh Vow that was easy....
Thanks Banerjee
Anyways, If you don't mind I would like to ask you where did you study probability from??
I mean the source of prob questions apart from Gmat Club?
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[#permalink]  02 Mar 2005, 17:37
sometime back I reviewed some rules.....in comb, pert and prob from this club, nothing other than that. I like to think of these as questions in which all u have to do is to figure out the steps u need to reach a condition i.e. they can be steps involving "and" or may be "or".....paraphrase ur ques in ur head and see what u ought to do to reach the condition in the ques and know the "and", "or" rules...i.e. multiply when "and" (only mutually independent) and add when u see an "or" when u paraphrase. U will see that u will never need to mug any formulas for any of these questions. Try this with the ques above and see if it works for u.
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[#permalink]  03 Mar 2005, 03:29
HongHu wrote:
Good job baner!

christoph wrote:
hmm ? what is wrong with these answers ?

1. 2 men and 4 women = 8c2*8c4 and 3 men and 3 women = 8c3 * 8c3 and 4 men and 2 women = 8c4 * 8c2 => add them

2. 7c3 (without robert) * 8c3 (with ann) + 7c3 (without ann) * 8c3 (with robert)

I believe your solution for 1 is correct. But you've got to picked up a lot repetitions for your second solution. The first 7C3*8C3 have included outcomes that have Ann and don't have Ann. And the second 7C3*8C3 have included outcomes that have Robert and don't have Robert. Does this make sense to you? Perhaps you could calculate the numbers out to compare the answers from the different approaches. When I have time in my test, I'd do two approaches to verify my answer too.

the solution as a result of baners calculation is:

1. 51744 and 2. 2695

the solution as a result of my calculation is:

1. 7056 <=> where am i wrong ? and 2. you are right there are indeed to many repetitions. i would prefer baners approach.
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[#permalink]  03 Mar 2005, 09:06
HongHu wrote:
Want to make it a little harder?;)

2. A club has 8 male including Robert and 8 female members including Ann. The club is choosing a commitee of 6 members. The commitee must have 3 male and 3 female members. Robert and Ann refused to serve in the same committee. How many different commitees must be chosen ?

C(8,3)*C(8,3) - C(7,2)*C(7,2)

OH I THINK MY APPROACH FOR THIS PART IS SAME AS BANERJEE'S.. I AGREE WITH THIS..... ANN AND ROBERT ARE ASSUMED TO BE IN THE COMMITTEE AND THUS WE ARE LEFT WITH 2 MALES AND 2 FEMALES MEMEBERS TO BE CHOSEN OUT OF 7 MALES AND 7 FEMALES RESPECTIVELY....
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[#permalink]  03 Mar 2005, 10:31
Banner is the man.

Btw guys, I've shortened some of the quotes out from your posts. If your post follows the one you are answering immediately the quotes may not be necessary. It makes it easier for others to read. (Plus we are having quite frequent server problems due to heavy load I thought this may (or may not) reduce the load a little?)
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[#permalink]  03 Mar 2005, 10:52
christoph wrote:
HongHu wrote:

christoph wrote:
hmm ? what is wrong with these answers ?

1. 2 men and 4 women = 8c2*8c4 and 3 men and 3 women = 8c3 * 8c3 and 4 men and 2 women = 8c4 * 8c2 => add them

I believe your solution for 1 is correct.

the solution as a result of baner`s calculation is:

51744
the solution as a result of my calculation is:

1. 7056 <=> where am i wrong ?

Actually, I now believe your appoach is right and banner is wrong. Just think about this: Why C(8,4) is not equal to C(8,2)*C(6,2)?
SVP
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[#permalink]  06 Mar 2005, 20:48
Nobody had given this one any more thoughts?
Senior Manager
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[#permalink]  07 Mar 2005, 08:00
HongHu wrote:
Nobody had given this one any more thoughts?

HongHu, can you pls post the final answer with a good explanation. Thanks;)
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[#permalink]  07 Mar 2005, 08:23
Ok let's see. The question is:

Quote:
1. A club has 8 male and 8 female members. The club is choosing a commitee of 6 members. The commitee must have at least 2 male and 2 female members. How many different commitees must be chosen ?

Approach 1:
First choose 2 male from the 8 male: C(8,2)
Then choose 2 female from the 8 female: C(8,2)
Then choose the 2 remaining from the 12 remaining: C(12,2)
Total outcome = C(8,2)*C(8,2)*C(12,2) = (8*7/2)*(8*7/2)*(12*11/2) = (8*7/2)*(8*7/2)*(66)

Approach 2:
Total outcome for choosing 2 male and 4 female: C(8,2)*C(8,4)
Total outcome for choosing 3 male and 3 female: C(8,3)*C(8,3)
Total outcome for choosing 4 male and 2 famale: C(8,4)*C(8,2)

Total outcome = C(8,2)*C(8,4)*2+C(8,3)*C(8,3)
= (8*7/2)*(8*7*6*5/4*3*2*1)*2+(8*7*6/3*2*1)*(8*7*6/3*2*1)
= (8*7/2)*(8*7/2)*[(6*5/4*3)*2 + (6/3)*(6/3)]
= (8*7/2)*(8*7/2)*(9)

Compare the answer for the two approaches, obviously 9 is different from 66. What is wrong?

The mistake of approach 1 is that it has included some cases multiple times. For example, say we chose male A and B, female C and D, and then we choose male E and female F from the rest of people. This is counted as one outcome. However, we could also chose male A and E, female C and D, and then we choose male B and female F from the rest of the people. It is counted as another outcome. On a closer look however, you'll notice that these two supposedly different outcomes are actually the same thing.

Normally we'd be able to get to the number of non-ordered outcomes from the number of ordered outcomes by dividing the first by the number of different orders. However in this case it is kind of hard to get the number of different orders. So I would recommend we use approach 2 to solve this question.

Last edited by HongHu on 12 Mar 2005, 21:31, edited 1 time in total.
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[#permalink]  07 Mar 2005, 08:37
banerjeea_98 wrote:
there is only 1 way to choose both Rob and Ann and then 7C2*7C2 ways to choose for the rest of the 4 positions.

wouldnt there be two ways to choose rob and ann first rob then ann and ann then rob?
so shouldnt it be 2*7C2*7C2?
[#permalink] 07 Mar 2005, 08:37

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