It is currently 20 Nov 2017, 01:06

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Director
Joined: 10 Feb 2006
Posts: 654

Kudos [?]: 634 [1], given: 0

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

### Show Tags

19 Nov 2007, 06:21
1
KUDOS
19
This post was
BOOKMARKED
00:00

Difficulty:

15% (low)

Question Stats:

74% (00:50) correct 26% (00:44) wrong based on 574 sessions

### HideShow timer Statistics

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

_________________

GMAT the final frontie!!!.

Kudos [?]: 634 [1], given: 0

Manager
Joined: 26 Sep 2007
Posts: 65

Kudos [?]: 28 [9], given: 5

### Show Tags

21 Nov 2007, 08:48
9
KUDOS
3
This post was
BOOKMARKED
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

Kudos [?]: 28 [9], given: 5

SVP
Joined: 28 Dec 2005
Posts: 1543

Kudos [?]: 185 [6], given: 2

### Show Tags

20 Nov 2007, 20:25
6
KUDOS
2
This post was
BOOKMARKED
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

Kudos [?]: 185 [6], given: 2

Current Student
Joined: 02 Apr 2012
Posts: 77

Kudos [?]: 60 [5], given: 155

Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
WE: Consulting (Consulting)
Re: A certain law firm consists of 4 senior partners and 6 [#permalink]

### Show Tags

17 Jul 2013, 13:49
5
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
_________________

Encourage cooperation! If this post was very useful, kudos are welcome
"It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

Kudos [?]: 60 [5], given: 155

Math Expert
Joined: 02 Sep 2009
Posts: 42259

Kudos [?]: 132744 [4], given: 12371

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

### Show Tags

17 Jul 2013, 14:49
4
KUDOS
Expert's post
11
This post was
BOOKMARKED
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
_________________

Kudos [?]: 132744 [4], given: 12371

SVP
Joined: 24 Aug 2006
Posts: 2130

Kudos [?]: 154 [2], given: 0

### Show Tags

21 Feb 2008, 12:47
2
KUDOS
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.

Kudos [?]: 154 [2], given: 0

CEO
Joined: 21 Jan 2007
Posts: 2734

Kudos [?]: 1076 [1], given: 4

Location: New York City

### Show Tags

21 Feb 2008, 12:38
1
KUDOS
At least one = Total - none

At least one = 10C3 - none

None = 6C3

= 10C3 - 6C3
= 120 - 20 = 100
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

Kudos [?]: 1076 [1], given: 4

Intern
Joined: 02 Aug 2007
Posts: 36

Kudos [?]: 43 [0], given: 0

### Show Tags

19 Nov 2007, 08:18
2
This post was
BOOKMARKED
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

Kudos [?]: 43 [0], given: 0

Manager
Joined: 25 Jul 2007
Posts: 107

Kudos [?]: 22 [0], given: 0

### Show Tags

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.

Kudos [?]: 22 [0], given: 0

SVP
Joined: 28 Dec 2005
Posts: 1543

Kudos [?]: 185 [0], given: 2

### Show Tags

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 ?

Kudos [?]: 185 [0], given: 2

Senior Manager
Joined: 06 Aug 2007
Posts: 360

Kudos [?]: 35 [0], given: 0

### Show Tags

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.

Kudos [?]: 35 [0], given: 0

Intern
Joined: 27 May 2008
Posts: 1

Kudos [?]: [0], given: 0

### Show Tags

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...

Kudos [?]: [0], given: 0

Manager
Joined: 27 Oct 2008
Posts: 185

Kudos [?]: 166 [0], given: 3

### Show Tags

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

Kudos [?]: 166 [0], given: 3

Senior Manager
Joined: 22 Dec 2009
Posts: 356

Kudos [?]: 418 [0], given: 47

### Show Tags

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)
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

Kudos [?]: 418 [0], given: 47

Re: PS - Combination   [#permalink] 16 Feb 2010, 13:32
Display posts from previous: Sort by