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# A certain law firm consists of 4 senior partners and 6

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Director
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A certain law firm consists of 4 senior partners and 6 [#permalink]

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19 Nov 2007, 06:21
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Difficulty:

15% (low)

Question Stats:

75% (01:51) correct 25% (00:53) wrong based on 348 sessions

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A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (2 groups are considered different if at least one group member is different)

A. 48
B. 100
C. 120
D. 288
E. 600

My take :

4C1 * 6C2 + 4C2 *6C1

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-firm-has-4-senior-partners-and-6-junior-partners-how-many-106010.html
[Reveal] Spoiler: OA

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Re: A certain law firm consists of 4 senior partners and 6 [#permalink]

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17 Jul 2013, 14:49
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Maxirosario2012 wrote:
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed 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)

48
100
120
288
600

Another approach:

SSSS JJJJJJ

We have 4 seniors and 6 juniors.
We are asked for the nomber of groups of 3 in which at least 1 is a senior (1, 2 or 3 seniors in each group).
Considering that we have 3 types of groups:

Groups with 1 senior:
We take 1 senior out of 4 ($$C^4_1$$) and combine them with 2 juniors out of 6 ($$C^6_2$$):

$$C^4_1*C^6_2 = 4*15 = 60$$

Groups with 2 seniors:
We take 2 seniors out of4 ($$C^4_2$$) and combine them with 1 juniors out of 6 ($$C^6_1$$):

$$C^4_2*C^6_1 = 6*6 = 36$$

Groups with 3 seniors (and zero juniors):
We take 3 seniors out of 4 ($$C^4_3$$):

$$C^4_3$$ = 4

60 + 36 + 4 = 100

A firm has 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed in which at least one member of the group is a senior partner. (2 groups are considered different if at least one group member is different)
A. 48
B. 100
C. 120
D. 288
E. 600

Total # of different groups of 3 out of 10 people: $$C^3_{10}=120$$;
# of groups with only junior partners (so with zero senior member): $$C^3_6=20$$;

So the # of groups with at least one senior partner is {all} - {none}= {at least one} = 120-20 = 100.

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-firm-has-4-senior-partners-and-6-junior-partners-how-many-106010.html
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20 Nov 2007, 20:25
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GMATBLACKBELT wrote:
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed 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)

48
100
120
288
600

My take :

4C1 * 6C2 + 4C2 *6C1

im really lost on this problem...

What really helps me in these type of problems is to approach the thought process like this:

1. What are the total # of combinations without any restrictions ?

2. What are the total # of combinations using the OPPOSITE of the restriction ?

3. The # of combinations with the restriction is the difference between 1 and 2

So, for this question, 1 would be 10C3. 2 would be 6C3, which gives the # of ways to pick 3 ppl out of the 6 junior partners, i.e. no seniors.

The final answer should be 10C3 - 6C3
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21 Nov 2007, 08:48
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Total no of groups of 3 members (including junior and senior) = 10C3
Total no of groups of 3 members (only juniors) = 6C3
Total no of groups of 3 members (at least 1 senior) = 10C3 - 6C3 = 120 - 20 =100
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Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
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Re: A certain law firm consists of 4 senior partners and 6 [#permalink]

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17 Jul 2013, 13:49
3
KUDOS
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed 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)

48
100
120
288
600

Another approach:

SSSS JJJJJJ

We have 4 seniors and 6 juniors.
We are asked for the nomber of groups of 3 in which at least 1 is a senior (1, 2 or 3 seniors in each group).
Considering that we have 3 types of groups:

Groups with 1 senior:
We take 1 senior out of 4 ($$C^4_1$$) and combine them with 2 juniors out of 6 ($$C^6_2$$):

$$C^4_1*C^6_2 = 4*15 = 60$$

Groups with 2 seniors:
We take 2 seniors out of4 ($$C^4_2$$) and combine them with 1 juniors out of 6 ($$C^6_1$$):

$$C^4_2*C^6_1 = 6*6 = 36$$

Groups with 3 seniors (and zero juniors):
We take 3 seniors out of 4 ($$C^4_3$$):

$$C^4_3$$ = 4

60 + 36 + 4 = 100
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21 Feb 2008, 12:47
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All possibilities = 10!/7!3! = 120

So the answer will be a fewer than 120; 48 is of course too few since it's less than half, so 100 looks good enough for me.
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19 Nov 2007, 08:18
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed 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)

48
100
120
288
600

My take :

4C1 * 6C2 + 4C2 *6C1

1Senior2Juniors + 2Seniors1Junior + 3Seniors =
4C1 * 6C2 + 4C2 *6C1 + 4C3
Manager
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19 Nov 2007, 08:24
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed 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)

48
100
120
288
600

My take :

4C1 * 6C2 + 4C2 *6C1

Your method is correct but you have not considered the possibility of all the 3 members of the group being senior members.
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19 Nov 2007, 21:13

# of teams with at least 1 senior = total # of teams - # of teams with 0 senior partners

total # = 10C3

# of teams with 0 seniors = 6C3

but this didnt give me the right answer .... what am i missing ?
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24 Nov 2007, 00:26
pmenon wrote:
GMATBLACKBELT wrote:
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed 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)

48
100
120
288
600

My take :

4C1 * 6C2 + 4C2 *6C1

im really lost on this problem...

What really helps me in these type of problems is to approach the thought process like this:

1. What are the total # of combinations without any restrictions ?

2. What are the total # of combinations using the OPPOSITE of the restriction ?

3. The # of combinations with the restriction is the difference between 1 and 2

So, for this question, 1 would be 10C3. 2 would be 6C3, which gives the # of ways to pick 3 ppl out of the 6 junior partners, i.e. no seniors.

The final answer should be 10C3 - 6C3

pmenon great explanation!! I totally lost this one.
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21 Feb 2008, 12:38
At least one = Total - none

At least one = 10C3 - none

None = 6C3

= 10C3 - 6C3
= 120 - 20 = 100
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05 Aug 2008, 01:06
But why 10C3 - 6C3?

Shouldn't it be 10P3 - 6P3

Bcoz the question mentions
(Two groups are considered different if at least one group member is different)

Am I missing something here...
Manager
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27 Sep 2009, 23:30
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed 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)

48
100
120
288
600

Soln:
= 4C2 * 6C1 + 4C1 *6C2 + 4C3
= 100 ways
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16 Feb 2010, 13:32
A certain law firm consists of 4 senior partners and 6 junior partners. How many different groups of 3 partners can be formed 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)

48
100
120
288
600

My take :

4C1 * 6C2 + 4C2 *6C1

Ans = Total Comb - Comb with only junior partners = 10c3 - 6c3 = 100 (B)
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Re: PS - Combination   [#permalink] 16 Feb 2010, 13:32
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