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tricky question from GMAT Prep - instructors!?

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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tricky question from GMAT Prep - instructors!?

by san2009 » Sat Jul 31, 2010 12:36 am
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
A) 48
b) 100
c) 120
d) 288
e) 600

Why doesn't the following approach work??

Since 1 member must be a senior partner
we are left with 9 individuals and the question becomes...how many different ways can we select 2 members from these 9 individuals

this can be represented as:
4 * (9!/(2!*7!)) = 4*36=144
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by outreach » Sat Jul 31, 2010 1:29 am
it is atleast one member and not only 1
is the OA correct..i think the correct ans is 100
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by outreach » Sat Jul 31, 2010 1:42 am
100 is correct

https://www.manhattangmat.com/forums/a-c ... t4110.html
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by san2009 » Sat Jul 31, 2010 1:45 am
that doesnt answer my question.
there are two other methods to solve this...which i understand but
what im trying to understand is why my method above is not working.
thanks
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by paruloberai » Sat Jul 31, 2010 8:45 am
san2009 wrote:A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
A) 48
b) 100
c) 120
d) 288
e) 600

Why doesn't the following approach work??

Since 1 member must be a senior partner
we are left with 9 individuals and the question becomes...how many different ways can we select 2 members from these 9 individuals

this can be represented as:
4 * (9!/(2!*7!)) = 4*36=144

# of ways to get 1 senior partner and 2 junior partners: (4! / 1!3!) x (6! / 2!4!) = 4 x 15 = 60
# of ways to get 2 senior partners and 1 junior partner: (4! / 2!2!) x (6! / 1!5!) = 6 x 6 = 36
# of ways to get 3 senior partners and 0 junior partners: (4! / 3!1!) x (6! / 0!6!) = 4 x 1 = 4
add these up = 100
Parul Oberai is a content expert for GMATLounge. She has a lot of experience helping GMAT students to be more efficient in solving quantitative problems. She can be reached at https://gmatlounge.com/ .
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by paruloberai » Sat Jul 31, 2010 8:48 am
san2009 wrote:A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
A) 48
b) 100
c) 120
d) 288
e) 600

Why doesn't the following approach work??

Since 1 member must be a senior partner
we are left with 9 individuals and the question becomes...how many different ways can we select 2 members from these 9 individuals

this can be represented as:
4 * (9!/(2!*7!)) = 4*36=144

# of ways to get 1 senior partner and 2 junior partners: (4! / 1!3!) x (6! / 2!4!) = 4 x 15 = 60
# of ways to get 2 senior partners and 1 junior partner: (4! / 2!2!) x (6! / 1!5!) = 6 x 6 = 36
# of ways to get 3 senior partners and 0 junior partners: (4! / 3!1!) x (6! / 0!6!) = 4 x 1 = 4
add these up = 100
Parul Oberai is a content expert for GMATLounge. She has a lot of experience helping GMAT students to be more efficient in solving quantitative problems. She can be reached at https://gmatlounge.com/ .
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by san2009 » Sat Jul 31, 2010 9:20 am
thanks. i am aware of that method but im interested in finding out what's wrong with the approach i outlines
thx
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by nithi_mystics » Mon Aug 02, 2010 7:40 am
According to your approach,

Senior1 , Senior2 and Senior3 is one group and
Senior2 , Senior1 and Senior3 is another group.

That's the mistake in your approach.
san2009 wrote:A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formd in which at least one member of the group is a senior partner? (Two groups are considered different if at least one group member is different)
A) 48
b) 100
c) 120
d) 288
e) 600

Why doesn't the following approach work??

Since 1 member must be a senior partner
we are left with 9 individuals and the question becomes...how many different ways can we select 2 members from these 9 individuals

this can be represented as:
4 * (9!/(2!*7!)) = 4*36=144
Thanks
Nithi
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by bdevas01 » Mon Aug 02, 2010 9:01 am
This is the best way to solve this problem:

Set up an equation:

A - B = C

A = Total possible number of groups
B = Total number of groups without senior partners
C = Total number of groups with at least one senior partner

Now solve for each variable using combinatorics:

A = 10 C 3, which is 10! / (3!7!) = 120

So there are 120 total ways of choosing 3 partners out of the 10.

B = 6 C 3, which equals 10! / (3!3!) = 20

Which means we have 20 ways of making 3-partner teams out of the 6 junior partners.

Now solve for C:

A - B = C

120 - 20 = 100

So there are 100 total ways of choosing 3-partner teams with at least one senior partner.
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