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Manager  Joined: 22 Jul 2009
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A firm is divided into four departments, each of which contains four  [#permalink]

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Question Stats: 63% (01:41) correct 37% (02:03) wrong based on 177 sessions

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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

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

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helloanupam 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)

I did receive an explanation but did not really understand. My way to comprehend this was as follows,

Since there are 3 people to be choosen (and none can be repeated from the same department);
the number of ways to choose the departments are 4C3 ways. For selecting a person from each department is 4C1 X 4C1 X 4C1 ways. So the total number of ways (I presume my calculation is leading to distinct teams) is
4C3 X 4C1X 4C1 X4C1 ways= 4^4 ways. Could anyone please validate my thinking or correct me. Thanks.

Yes, your solution is correct.

Basically we are asked to determine the # of different groups of 3 that can be formed so that no 2 members are from the same department.

$$C^3_4=4$$ - choosing which 3 departments out of 4 will provide with a member;
$$C^1_4*C^1_4*C^1_4=4^3$$ - choosing each member from the selected 3 departments;

So, total # of different groups will be: $$4*4^3=4^4$$.

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Manager  Joined: 18 Jul 2009
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Schools: South Asian B-schools
Re: A firm is divided into four departments, each of which contains four  [#permalink]

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

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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

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.
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Intern  Joined: 06 Jul 2010
Posts: 14
Re: A firm is divided into four departments, each of which contains four  [#permalink]

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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

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

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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.

What if we had six grps, wud the answer be 6C3 * 4*4 *4 ? GMAT experts, please help.
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Re: A firm is divided into four departments, each of which contains four  [#permalink]

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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.

What if we had six grps, wud the answer be 6C3 * 4*4 *4 ? GMAT experts, please help.

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

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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)

I did receive an explanation but did not really understand. My way to comprehend this was as follows,

Since there are 3 people to be choosen (and none can be repeated from the same department);
the number of ways to choose the departments are 4C3 ways. For selecting a person from each department is 4C1 X 4C1 X 4C1 ways. So the total number of ways (I presume my calculation is leading to distinct teams) is
4C3 X 4C1X 4C1 X4C1 ways= 4^4 ways. Could anyone please validate my thinking or correct me. Thanks.
Intern  Joined: 14 May 2011
Posts: 5
Re: A firm is divided into four departments, each of which contains four  [#permalink]

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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
Retired Moderator Joined: 20 Dec 2010
Posts: 1588
Re: A firm is divided into four departments, each of which contains four  [#permalink]

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

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I'm still confused. How would this question be answered if I was using the slotting method?
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Posts: 58335
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pitpat wrote:
I'm still confused. How would this question be answered if I was using the slotting method?

First member: 4*4=16 choices;
Second member: 4*3=12 choices (as one department already provided with a member);
Third member: 4*2=8 choices (as two departments already provided with members);

So, we have: 16*12*8.

Since the order of the members in a team does not matter (we don't actually have 1st, 2nd and 3rd members) then we should divide above by 3! to get rid of duplication: 16*12*8/3!=4^4.

As for the other approach. Consider this: $$C^3_4$$ is # of ways to choose which 3 departments will provide with an employee from 4 departments for the team. Now, since 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's clear.
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