It is currently 23 Jun 2017, 10:55

### 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 firm has 4 senior partners and 6 junior partners. How many

Author Message
TAGS:

### Hide Tags

Intern
Joined: 20 Oct 2010
Posts: 10
A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

09 Dec 2010, 08:21
1
KUDOS
8
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

69% (02:20) correct 31% (12:15) wrong based on 376 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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 39609
A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

09 Dec 2010, 09:38
2
KUDOS
Expert's post
6
This post was
BOOKMARKED
gdk800 wrote:
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 memeber): $$C^3_6=20$$;

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

_________________
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 9254
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: 340 Q170 V170
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

09 Oct 2016, 22:04
1
KUDOS
Expert's post
Hi ashwinchivukula,

Unfortunately, your math includes some 'duplicate entries.'

For example, let's call the 4 senior partners A, B, C and D and the 6 junior partners 1, 2, 3, 4, 5 and 6.

In your calculation, you state that the first person selected MUST be one of those 4 seniors (A/B/C/D) and the remaining two people can be any two of the remaining 9...

The group "A/B/1" and "B/A/1" are the SAME group, but your calculation counts THAT group TWICE (depending on whether A or B was chosen first). In a Combination question, you can't allow duplicate entries.

GMAT assassins aren't born, they're made,
Rich
_________________

760+: Learn What GMAT Assassins Do to Score at the Highest Levels
Contact Rich at: Rich.C@empowergmat.com

# Rich Cohen

Co-Founder & GMAT Assassin

# Special Offer: Save $75 + GMAT Club Tests 60-point improvement guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Math Expert Joined: 02 Sep 2009 Posts: 39609 Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 20 May 2017, 05:32 1 This post received KUDOS Expert's post rocko911 wrote: Bunuel wrote: gdk800 wrote: 1) 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 [Reveal] Spoiler: OA is B. Total # of different groups of 3 out of 10 people: $$C^3_{10}=120$$; # of groups with only junior partners (so with zero senior memeber): $$C^3_6=20$$; So the # of groups with at least one senior partner is {all} - {none}= {at least one} = 120-20 = 100. Answer: B. Your explanation says that there can be all 3 senior members also as mentioned AT LEAST 1 senior so there can be 3 at most , FINE But in the last line (2 groups are considered different if at least one group member is different), if we consider this , then it means we need atleast one member should be different in a group , So 4C1 * 6C2 (1 senior 2 junior) + 4C2 * 6C1 (2 senior and 1 junior) + 4C3 (Why this 4C3? As we need atleast one different member but in this 4C3 we will have all seniors and no junior and this will contradict the last line) Please help? 2 groups are considered different if at least one group member is different does NOT mean that the group must have members from both seniors and juniors. It means that (ABC) and (ABD) considered different groups because at least one group member is different, while (ABC) and (ACB) are NOT considered different because all members of these groups are the same even the order is different. This part (2 groups are considered different if at least one group member is different) is added there to hint that the order of the group is not important and only the members of the group are. Hope it's clear. _________________ Intern Joined: 14 Apr 2012 Posts: 1 senior partners [#permalink] ### Show Tags 20 Apr 2013, 11:22 What seems to work as well, though its tougher to come up with this solution: (10*9*8) * 2/3 + (10*9*4)*1/3 = 600. Divided by all possible permutations (=3!) yields 100. Explanation: For the first spot you have 10 possible canditates, for the 2nd 9. For the 3rd spot you need to differentiate if a senior partner has been picked for one of the first 2 spots or not. If yes, then you can pick one of the 8 remaining guys, so 10*9*8. If no senior partner has been picked yet, you need 10*9*4. Now you need to weight the 2 cases with the corresponding probabilities: probability of no senior partner being picked in the first 2 draws = 6/10*5/9=1/3, so you weight 10*9*4 with 1/3. For the complementary case (senior partner was picked in the first 2 draws) you just take the complementary prob (1-1/3)= 2/3 and weight 10*9*8 with it. Now you just need to divide the answer (600) by the number of different positions (=3!) and get 600/6=100 But I suggest you stick with the easy solution in the GMAT Current Student Joined: 04 Mar 2013 Posts: 69 Location: India Concentration: Strategy, Operations Schools: Booth '17 (M) GMAT 1: 770 Q50 V44 GPA: 3.66 WE: Operations (Manufacturing) Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 20 Apr 2013, 18:57 The standard approach 1 senior partner 4 x 6C2 = 60 2 senior partners 4C2 x 6 = 36 3 senior partners 4C3 = 4 Total 100 However GDKs approach is a better way around _________________ When you feel like giving up, remember why you held on for so long in the first place. Director Joined: 17 Dec 2012 Posts: 550 Location: India Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 22 Apr 2013, 22:44 gdk800 wrote: 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 A general approach to solving Combination problems Steps: 1. There are two larger groups, the senior partners and the junior partners. 2. The first larger group i.e., the senior partners is 4 in number. So let$$n1$$ be 4. The second larger group, ie, the junior partners is 6 in number. So $$n2$$ is 6. 3. The smaller group that is selected from the larger group of senior partners may be 1, 2, or 3 in number. So $$r1$$ is 1 or 2 or 3 . Correspondingly the other smaller group i.e.,$$r2$$ selected from the junior partners is 2 or 1 or 0 in number. 4. For each value of$$r1$$ and the corresponding$$r2$$, compute the number of combinations which are$$4C1 * 6C2$$, $$4C2 * 6C1$$ and$$4C3 * 6C0$$ being 60, 36 and 4 ways respectively. 5. The total number of combinations is therefore 60+36+4 = 100 _________________ Srinivasan Vaidyaraman Sravna http://www.sravnatestprep.com Classroom and Online Coaching Manager Joined: 04 Oct 2013 Posts: 177 Concentration: Finance, Leadership GMAT 1: 590 Q40 V30 GMAT 2: 730 Q49 V40 WE: Project Management (Entertainment and Sports) A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 28 Dec 2014, 08:49 You can also think about it as a code. 4S - 6J three different ways to group them: S J J. 4(6)5/2! because we have two junior members or S S J. 4(3)6/2! because we have two senior members or S S S 4(3)2/3! because we have three senior members. add up the results and you get 100. _________________ learn the rules of the game, then play better than anyone else. Last edited by gmat6nplus1 on 28 Dec 2014, 10:38, edited 1 time in total. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 9254 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink] ### Show Tags 28 Dec 2014, 10:07 Hi All, This is a rare layered Combinatorics question. To answer it, you can either figure out the Total of ALL possibilities and subject the ones that "don't fit" or you can perform 3 separate calculations to account for the ones that "do fit." Here is how you can approach the latter option: We're given 4 senior partners and 6 junior partners. We're asked for the number of different groups of 3 (the clue that we'll need the Combination Formula) with one stipulation - there must be AT LEAST 1 senior partner. Here are the 3 calculations: 1) 3 senior partners = 4c3 = 4!/[3!1!] = 4 different groups 2) 2 seniors and 1 junior = (4c2)(6c1) = (4!/[2!2!])(6!/[1!5!) = (6)(6) = 36 different groups 3) 1 senior and 2 juniors = (4c1)(6c2) = (4!/[3!1!])(6!/[2!4!]) = (4)(15) = 60 different groups 4 + 36 + 60 = 100 different groups GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin # Special Offer: Save$75 + GMAT Club Tests

60-point improvement guarantee
www.empowergmat.com/

***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***********************

VP
Joined: 09 Jun 2010
Posts: 1413
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

10 May 2015, 02:40
Bunuel wrote:
gdk800 wrote:
1) 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

[Reveal] Spoiler:
OA is B.

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

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

Princeton book is wonderful for explanation of coounting and combination. read it to understand of the math legendary
_________________

visit my facebook to help me.
on facebook, my name is: thang thang thang

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 15926
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

24 Jun 2016, 12:34
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Director
Joined: 18 Oct 2014
Posts: 908
Location: United States
GMAT 1: 660 Q49 V31
GPA: 3.98
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

24 Jun 2016, 14:02
gdk800 wrote:
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

Ways to choose at least 1 Senior

1 S and 2 J 4C1*6C2= 60
2 S and 1 J 4C2 * 6C1= 36
3 S and 0 J 4C3 = 4

60+36+4= 100

_________________

I welcome critical analysis of my post!! That will help me reach 700+

Intern
Joined: 09 Oct 2016
Posts: 1
A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

09 Oct 2016, 20:24
Hi,

Can some one help me understand what is wrong with my logic here?

A group should have 3 partners with at least 1 senior partner.
Overall there are 4 senior partners & 6 junior partners.

At least 1 senior partner = 4C1
Remaining 2 can be anyone from the remaining partners = (10-1)C2

4C1*9C2 = 144?

Can some one help me understand what i am missing here?
Manager
Status: In the realms of Chaos & Night
Joined: 13 Sep 2015
Posts: 172
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

18 Dec 2016, 05:29
Bunuel wrote:
gdk800 wrote:
1) 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

[Reveal] Spoiler:
OA is B.

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

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

Hi Bunuel

The question says - "How many different groups of 3 partners can be formed "
So we can calculate as:

Case - I
One group of 3Sp + 1Jp + second group of 1Sp + 2Jp

Case - II
One Group of 2Sp + 2Jp + Second group of 2Sp + 2Jp

Case -III
One Group of 2Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

Case - IV
One Group of 1Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

As per the ans - Its how many way a group of 3 Partners can be formed.
Or I seem to misinterpret the prompt.
_________________

Good luck
=========================================================================================
"If a street performer makes you stop walking, you owe him a buck"
"If this post helps you on your GMAT journey, drop a +1 Kudo "

"Thursdays with Ron - Consolidated Verbal Master List - Updated"

Math Expert
Joined: 02 Sep 2009
Posts: 39609
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

18 Dec 2016, 23:09
Nightfury14 wrote:
Bunuel wrote:
gdk800 wrote:
1) 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

[Reveal] Spoiler:
OA is B.

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

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

Hi Bunuel

The question says - "How many different groups of 3 partners can be formed "
So we can calculate as:

Case - I
One group of 3Sp + 1Jp + second group of 1Sp + 2Jp

Case - II
One Group of 2Sp + 2Jp + Second group of 2Sp + 2Jp

Case -III
One Group of 2Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

Case - IV
One Group of 1Sp + 2Jp + Second group of 1Sp + 2Jp + Third Group of 1Sp + 2Jp

As per the ans - Its how many way a group of 3 Partners can be formed.
Or I seem to misinterpret the prompt.

The group must have 3 partners in it out of which at least one member is a senior partner:
3SP
2SP + 1JP
1SP + 2JP.
_________________
Intern
Joined: 18 May 2016
Posts: 20
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

18 Jan 2017, 17:21
Hello,

If one finds 48 by doing (4)(9)(8)/3!, what has he actually computed? Why is it incorrect to think that i can pick from 4 senior at first -obligatory-, then my choice is open to senior and junior (9 then 8). Why is this incorrect?
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 1158
Location: United States (CA)
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

07 Mar 2017, 07:56
gdk800 wrote:
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

We are asked to find the number of groups with at least one senior partner.

“At least 1” means "one or more," so the group must have 1 or 2 or 3 senior partners.

Case 1: Exactly 1 senior partner

Recall that the group must have 3 partners. Therefore, in this case, we need to pick 1 senior partner from 4 senior partners and 2 junior partners from 6 junior partners. The number of ways this can be done is 4C1 x 6C2.

4C1 x 6C2 = 4 x (6x5)/2! = 4 x 15 = 60

Case 2: Exactly 2 senior partners

In this case, we need to pick 2 senior partners from 4 senior partners and 1 junior partner from 6 junior partners. The number of ways this can be done is 4C2 x 6C1.

4C2 x 6C1 = (4x3)/2! x 6 = 6 x 6 = 36

Case 3: Exactly 3 senior partners

In this case, we need to pick 3 senior partners from 4 senior partners and no junior partners from 6 junior partners. The number of ways this can be done is 4C3 x 6C0.

4C3 x 6C0 = (4x3x2)/3! x 1 = 4 x 1 = 4

Thus, the total number of ways to form a group in which there is at least 1 senior partner = 60 + 36 + 4 = 100.

Alternate Solution:

It must be true that:

The total number of ways to form a group of 3 partners = (The number of ways in which the group would have at least 1 senior partner) + (The number of ways in which the group would have no senior partners).

Therefore:

The number of ways in which the group would have at least 1 senior partner = (The total number of ways to form a group of 3 partners) - (The number of ways in which the group would have no senior partners).

If the group of 3 has all junior partners, and there are 6 junior partners total, then the group of all junior partners can be made in 6C3 ways.

6C3 = (6 x 5 x 4)/3! = 5 x 4 = 20

The total number of groups of 3 that can be formed from 10 partners is 10C3.

10C3 = (10 x 9 x 8)/3! = 5 x 3 x 8 = 120

Thus, the number of ways to form a group of 3 in which there is at least 1 senior partner = 120 - 20 = 100 ways.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Intern
Joined: 11 Feb 2017
Posts: 28
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

20 May 2017, 03:59
Bunuel wrote:
gdk800 wrote:
1) 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

[Reveal] Spoiler:
OA is B.

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

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

Your explanation says that there can be all 3 senior members also as mentioned AT LEAST 1 senior so there can be 3 at most , FINE

But in the last line (2 groups are considered different if at least one group member is different), if we consider this , then it means we need atleast one member should be different in a group ,

So 4C1 * 6C2 (1 senior 2 junior) + 4C2 * 6C1 (2 senior and 1 junior) + 4C3 (Why this 4C3? As we need atleast one different member but in this 4C3 we will have all seniors and no junior and this will contradict the last line)

Director
Joined: 17 Dec 2012
Posts: 550
Location: India
Re: A firm has 4 senior partners and 6 junior partners. How many [#permalink]

### Show Tags

20 May 2017, 19:04
Expert's post
Top Contributor
gdk800 wrote:
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

1. It is a combination problem, because we need to select r out of n
2. Is there a constraint? There is a constraint. At least one of the member has to be a senior partner.
3. Since it is an "Atleast one" constraint, the opposite of the constraint "none" is easier to find.
4. Find the total number of combinations without constraint. It is 10C3.
5. Number of combinations where none is a senior partner is selecting 3 out of 6 junior partners= 6C3
6. Combinations with constraints is (4)-(5)=100
_________________

Srinivasan Vaidyaraman
Sravna
http://www.sravnatestprep.com

Classroom and Online Coaching

Re: A firm has 4 senior partners and 6 junior partners. How many   [#permalink] 20 May 2017, 19:04
Similar topics Replies Last post
Similar
Topics:
1/4 of all the juniors and 2/3 of all the seniors are going on a trip. 1 18 Sep 2016, 19:31
16 If 5 - 6/x = x, then x has how many possible values? 17 14 May 2017, 03:18
6 A firm is comprised of partners and associates in a ratio of 3 14 Dec 2014, 21:33
11 A certain company employs 6 senior officers and 4 junior of 6 30 Aug 2016, 17:38
51 A certain law firm consists of 4 senior partners and 6 13 17 Jul 2013, 14:49
Display posts from previous: Sort by