It is currently 18 Jan 2018, 21:56

### 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 is divided into four departments, each of which contains four

Author Message
TAGS:

### Hide Tags

Manager
Joined: 22 Jul 2009
Posts: 191

Kudos [?]: 292 [1], given: 18

A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

21 Sep 2009, 08:21
1
KUDOS
6
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

62% (01:18) correct 38% (01:25) wrong based on 228 sessions

### HideShow timer Statistics

A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?

(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

Source: Jeff Sackmann's GMAT Extreme Challenge
[Reveal] Spoiler: OA

_________________

Please kudos if my post helps.

Kudos [?]: 292 [1], given: 18

Manager
Joined: 18 Jul 2009
Posts: 167

Kudos [?]: 121 [1], given: 37

Location: India
Schools: South Asian B-schools
Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

21 Sep 2009, 09:08
1
KUDOS
my approach

4 Dept (I,II,III,IV) with 4 people in each dept.

To choose 1st team member = we have 16 options (4 dept x 4 people)
To choose 2nd member = we have 12 options (3 dept x 4 people)
To choose 3rd member = we have 8 options ( 2 dept x 4 people)

unique teams = 16 x 12 x 8...i don't know where i am wrong ..help me ...

_________________

Bhushan S.
If you like my post....Consider it for Kudos

Kudos [?]: 121 [1], given: 37

GMAT Tutor
Joined: 24 Jun 2008
Posts: 1347

Kudos [?]: 2082 [3], given: 6

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

21 Sep 2009, 11:30
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
powerka wrote:
A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three
people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?
(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

Source: Jeff Sackmann's GMAT Extreme Challenge

[Reveal] Spoiler:
B

First, we need to choose the three teams we'll take employees from. That's the same as choosing one team *not* to take an employee from, so we can choose our three teams in 4 ways. Then we have 4 choices for which employee we take from each of the three teams, so the answer is 4*4*4*4 = 4^4.

bhushan252 wrote:
my approach

4 Dept (I,II,III,IV) with 4 people in each dept.

To choose 1st team member = we have 16 options (4 dept x 4 people)
To choose 2nd member = we have 12 options (3 dept x 4 people)
To choose 3rd member = we have 8 options ( 2 dept x 4 people)

unique teams = 16 x 12 x 8...i don't know where i am wrong ..help me ...

You're assuming that there is a '1st team member', a '2nd team member' and a '3rd team member' - that is, you're assuming the order of the team members is somehow important. If the question asked in how many ways we might choose a President, Vice-President and Treasurer with the given restrictions, your answer would be correct. However, since the order of our 3 team members doesn't matter, you need to divide your answer by 3! = 6. That gives (16)(12)(8)/(6) = 16*2*8 = 2^8 = 4^4.
_________________

GMAT Tutor in Toronto

If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Kudos [?]: 2082 [3], given: 6

Manager
Joined: 22 Jul 2009
Posts: 191

Kudos [?]: 292 [0], given: 18

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

21 Sep 2009, 19:24
IanStewart wrote:
powerka wrote:
A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three
people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?
(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

Source: Jeff Sackmann's GMAT Extreme Challenge

[Reveal] Spoiler:
B

First, we need to choose the three teams we'll take employees from. That's the same as choosing one team *not* to take an employee from, so we can choose our three teams in 4 ways. Then we have 4 choices for which employee we take from each of the three teams, so the answer is 4*4*4*4 = 4^4.

bhushan252 wrote:
my approach

4 Dept (I,II,III,IV) with 4 people in each dept.

To choose 1st team member = we have 16 options (4 dept x 4 people)
To choose 2nd member = we have 12 options (3 dept x 4 people)
To choose 3rd member = we have 8 options ( 2 dept x 4 people)

unique teams = 16 x 12 x 8...i don't know where i am wrong ..help me ...

You're assuming that there is a '1st team member', a '2nd team member' and a '3rd team member' - that is, you're assuming the order of the team members is somehow important. If the question asked in how many ways we might choose a President, Vice-President and Treasurer with the given restrictions, your answer would be correct. However, since the order of our 3 team members doesn't matter, you need to divide your answer by 3! = 6. That gives (16)(12)(8)/(6) = 16*2*8 = 2^8 = 4^4.

Perfect explanation, thank you.

My approach was 4*4*4*4 = 4^4.
_________________

Please kudos if my post helps.

Kudos [?]: 292 [0], given: 18

Intern
Joined: 06 Jul 2010
Posts: 18

Kudos [?]: 5 [1], given: 1

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

07 Jul 2010, 19:16
1
KUDOS

Solution:
Each department has 4 people and we must select 1 person from each department: 4C1

There are 4 departments and we must form team of 3 people and no two person in the team must be from the same department: 4C1 * 4C1 * 4C1 * 4C0

Number of ways it can be arranged: 4

4*(4C1 * 4C1 * 4C1 * 4C0) = 4*4^3 = 4^4

Cheers
/Milap

Kudos [?]: 5 [1], given: 1

Manager
Joined: 04 Apr 2010
Posts: 158

Kudos [?]: 245 [1], given: 31

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

12 Jul 2010, 15:56
1
KUDOS
1
This post was
BOOKMARKED
@bhushan252
YOu calcualted the permutation, in which we consider a team (r, s, d) and (d,s,r) different. However they are same when it comes about selecting people.
SO divide you permutation result by r!
Permutation: n!/(n-r)!
Combintaiton: permutation/r!
so your result that is 16 x 12 x 8 divided by 3! is the answer that is 4^4 (B).
Cheers!
_________________

Consider me giving KUDOS, if you find my post helpful.
If at first you don't succeed, you're running about average. ~Anonymous

Kudos [?]: 245 [1], given: 31

Intern
Joined: 10 Jun 2010
Posts: 44

Kudos [?]: 51 [1], given: 2

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

13 Jul 2010, 04:30
1
KUDOS
Guys, I got the right answer, but i am not sure if my approach is right. Lets say we have only 3 groups from which we have to pick one person from each group. The the number of combinations would be 4*4*4. Now, since we have 4 groups and we need to pick three at a time, the number of ways we can do this is 4C1=4. Hence total number of ways u can pick 3 people (1 person from each grp) from four groups is 4 (4*4*4) = 4 ^ 4.

Kudos [?]: 51 [1], given: 2

Math Expert
Joined: 02 Sep 2009
Posts: 43322

Kudos [?]: 139451 [2], given: 12790

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

13 Jul 2010, 06:59
2
KUDOS
Expert's post
anandnat wrote:
Guys, I got the right answer, but i am not sure if my approach is right. Lets say we have only 3 groups from which we have to pick one person from each group. The the number of combinations would be 4*4*4. Now, since we have 4 groups and we need to pick three at a time, the number of ways we can do this is 4C1=4. Hence total number of ways u can pick 3 people (1 person from each grp) from four groups is 4 (4*4*4) = 4 ^ 4.

Correct. If there were 6 departments then: $$C^3_6$$ # of ways to choose which 3 department will provide employee for the team and as each chosen department can provide with 4 employees then total # of different teams will be $$C^3_6*4*4*4=5*4^4$$.

A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?
(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

The same approach for the original question: $$C^3_4$$ # of ways to choose which 3 department will provide employee for the team and as each chosen department can provide with 4 employees then total # of different teams will be $$C^3_4*4*4*4=4^4$$.

Hope it helps.
_________________

Kudos [?]: 139451 [2], given: 12790

Manager
Status: Waiting to hear from University of Texas at Austin
Joined: 24 May 2010
Posts: 76

Kudos [?]: 76 [0], given: 4

Location: Changchun, China
Schools: University of Texas at Austin, Michigan State
Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

14 Jul 2010, 20:03
First, when I read other peoples posted answers I think either I am thinking way too much here....or other people are not writing out all the steps.

The way I approached this problems is first I found the number of possible teams.

Since we have no replacements and order is not important I know it is combination

$$C^n_k$$

I have 16 people to choose from and I will only pick 3 of them

$$C^{16}_3$$

To add the condition no 2 people from the same department.

First Pick
Anyone

Second Pick
Any of 12, out of the remaining 15 = $$\frac{12}{15}$$

Third Pick
Any of 8, out of 14 remaining = $$\frac{8}{14}$$

So my total final equation to solve was $$C^{16}_3*\frac{16}{16}*\frac{12}{15}*\frac{8}{14}$$

I end up with the correct answer, but am I doing something wrong?

Kudos [?]: 76 [0], given: 4

Intern
Joined: 14 May 2011
Posts: 10

Kudos [?]: 7 [0], given: 2

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

31 Aug 2011, 06:02
I'm trying to solve the problem using reverse combination approach.Please,Can anyone help.

Need to choose 3 people from 16 so 16C3.

No. of ways to choose 3 people from the same team 4C3.since there are four teams (4C3)^4

16C3 - (4C3)^4

=560-256

=304

Kudos [?]: 7 [0], given: 2

Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1946

Kudos [?]: 2175 [1], given: 376

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

31 Aug 2011, 08:34
1
KUDOS
klueless7825 wrote:
I'm trying to solve the problem using reverse combination approach.Please,Can anyone help.

Need to choose 3 people from 16 so 16C3.

No. of ways to choose 3 people from the same team 4C3.since there are four teams (4C3)^4

16C3 - (4C3)^4

=560-256

=304

First of all, this approach is complicated, convoluted and error prone. I'd go by Bunuel's explanation. Different permutation and combination problems are done using different methods. You just need to pick the most apt one for a specific type.

Anyway,
No. of ways to choose 3 people from the same team 4C3.since there are four teams (4C3)^4

This is not entirely correct.
Number of ways to choose 3 people from the same team: 4C3. Correct
There are four teams: (4C3)^4. Not correct.

Remember, here you need to add the numbers not multiply because you choose from 1 OR choose from 2 OR choose from 3 OR choose from 4.

Secondly, you have missed out another case in which just 1 candidate is picked from 1 team AND 2 candidates are picked from another team. Because, that will also be the NOT one candidate from each team.

So, actual answer could be arrived by this:
$$Total=C^{16}_{3}=560$$

Picking all 3 candidates from just one team:
$$C^4_1*C^4_3=16$$
OR(i.e. +)
Picking 2 candidates from one team AND picking 1 candidate from one of the remaining teams
$$C^4_1*C^4_2*C^3_1*C^4_1=288$$

Total number of ways in which 1 candidate is NOT picked from each team
$$16+288=304$$

Now, to find the left-overs just subtract from total:
$$560-304=256=4^4$$

Ans: "B"
_________________

Kudos [?]: 2175 [1], given: 376

Intern
Joined: 14 May 2011
Posts: 10

Kudos [?]: 7 [0], given: 2

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

31 Aug 2011, 08:52
Thanks a lot for explaining the reverse combination approach.It helped me to understand combinations better.

Kudos [?]: 7 [0], given: 2

Senior Manager
Joined: 23 Oct 2010
Posts: 379

Kudos [?]: 418 [0], given: 73

Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

28 Sep 2011, 23:58
milapshah wrote:

Solution:
Each department has 4 people and we must select 1 person from each department: 4C1

There are 4 departments and we must form team of 3 people and no two person in the team must be from the same department: 4C1 * 4C1 * 4C1 * 4C0

Number of ways it can be arranged: 4

4*(4C1 * 4C1 * 4C1 * 4C0) = 4*4^3 = 4^4

Cheers
/Milap

I didnt get it. 4C1=4!/1*(4-1)!=4!/3! I mean 4C1 doesnt equal to 4
or I missed smth?
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Kudos [?]: 418 [0], given: 73

Senior Manager
Status: MBAing!!!!
Joined: 24 Jun 2011
Posts: 287

Kudos [?]: 52 [0], given: 56

Location: United States (FL)
Concentration: Finance, Real Estate
GPA: 3.65
WE: Project Management (Real Estate)
Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

30 Sep 2011, 00:33
Bunuel wrote:
anandnat wrote:
Guys, I got the right answer, but i am not sure if my approach is right. Lets say we have only 3 groups from which we have to pick one person from each group. The the number of combinations would be 4*4*4. Now, since we have 4 groups and we need to pick three at a time, the number of ways we can do this is 4C1=4. Hence total number of ways u can pick 3 people (1 person from each grp) from four groups is 4 (4*4*4) = 4 ^ 4.

Correct. If there were 6 departments then: $$C^3_6$$ # of ways to choose which 3 department will provide employee for the team and as each chosen department can provide with 4 employees then total # of different teams will be $$C^3_6*4*4*4=5*4^4$$.

A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?
(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

The same approach for the original question: $$C^3_4$$ # of ways to choose which 3 department will provide employee for the team and as each chosen department can provide with 4 employees then total # of different teams will be $$C^3_4*4*4*4=4^4$$.

Hope it helps.

Thanks Bunuel for the clear explanation. I got it.

Kudos [?]: 52 [0], given: 56

Intern
Joined: 09 Nov 2013
Posts: 17

Kudos [?]: 8 [0], given: 18

Location: United Arab Emirates
Concentration: Operations, Technology
Schools: MBS '16 (A)
GPA: 3.4
WE: Engineering (Energy and Utilities)
Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

12 Nov 2013, 07:12
This is the method that I followed.

Method 1:

You need to fill in three spots

_ x _ x _

from these guys
Group-1
(a,b,c,d)
Group-2
(e,f,g,h)
Group-3
(i,j,k,l)
Group-4
(m,n,o,p)

Since each member of the final 3 member group must be different, you choose one from each group (group-1,2,3 or 4). Now you have a choice of 4 people to select from, so for the group-1

4 x _ x _

group-2 again gives you four choices, so

4 x 4 x _

group-3 again gives you four choices

4 x 4 x 4

The same can be done using different combo of 4 groups. That combo is equal to 4C3 which is again equal to 4. Multiply that as well

4x4x4x4 = 4 ^ 4

Kudos [?]: 8 [0], given: 18

Current Student
Joined: 06 Sep 2013
Posts: 1957

Kudos [?]: 770 [0], given: 355

Concentration: Finance
Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

30 Dec 2013, 07:00
powerka wrote:
A firm is divided into four departments, each of which contains four people. If a project is to be assigned to a team of three people, none of which can be from the same department, what is the greatest number of distinct teams to which the project could be assigned?

(A) 4^3
(B) 4^4
(C) 4^5
(D) 6(4^4)
(E) 4(3^6)

Source: Jeff Sackmann's GMAT Extreme Challenge

We have four departments with 4 people in each

Now we need to pick 3 people of different departments, let's do it step by step

First, pick 3 departments among 4 different options
Hence 4C3 = 4

Now, pick between for people in each department
Hence, total combo is 4*4*4

Finally we then have 4^4 for total available options

B is the best choice here

Hope it helps!
Cheers!
J

PS. Does anybody have some more questions from Source: Jeff Sackmann's GMAT Extreme Challenge?

Will provide Kudos if you willl

Kudos [?]: 770 [0], given: 355

Intern
Joined: 12 Feb 2017
Posts: 1

Kudos [?]: 0 [0], given: 4

Re: A firm is divided into four departments, each of which contains four [#permalink]

### Show Tags

06 Jan 2018, 19:52
anandnat wrote:
Guys, I got the right answer, but i am not sure if my approach is right. Lets say we have only 3 groups from which we have to pick one person from each group. The the number of combinations would be 4*4*4. Now, since we have 4 groups and we need to pick three at a time, the number of ways we can do this is 4C1=4. Hence total number of ways u can pick 3 people (1 person from each grp) from four groups is 4 (4*4*4) = 4 ^ 4.

Actually, the correct representation of selecting 3 out of 4 groups is 4C3, which has the same value as 4C1 i.e. 4. It still gives you the same answer, since the values do not change.

Kudos [?]: 0 [0], given: 4

Re: A firm is divided into four departments, each of which contains four   [#permalink] 06 Jan 2018, 19:52
Display posts from previous: Sort by