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A firm is divided into four departments, each of which

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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)

[Reveal] Spoiler:
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.
[Reveal] Spoiler: OA
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New post 13 Dec 2010, 01:22
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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)

Answer: B.

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: \(4*4^3=4^4\).

Answer: B.
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Re: Combinations-4 Department,4 People-Assign to a team of 3. [#permalink]

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New post 13 Dec 2010, 01:58
Thanks! Bunuel.
Your explanations have always been lucid.
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New post 13 Dec 2010, 05:50
+1 for B

Thanks for explaining
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New post 17 Feb 2012, 00:15
I'm still confused. How would this question be answered if I was using the slotting method?
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New post 17 Feb 2012, 02:07
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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A firm is divided into four departments, each of which [#permalink]

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New post 09 Oct 2016, 10:02
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 was never good with combinatorics questions...yet this question can be solved by applying logic...

we have 16 people, out of whom we need to select 3.
16C3 = 16*15*14/6 -> 8*5*14 => 560.
since this gives the whole # of possibilities, in which even members of the same group are counted...it must be true that the total number must be less than 560...
1. 4^3 = 64 - too little
2. 4^4 = 256 - more real
3. 4^5 = 1024 - way too big
4. 4^4 = 256. multiply by 6 -> way too big.
5. 3^6 -> 3^4 = 81. 81*9 -> 729 -> way too big, and this is not even multiplying by 4.

we are left with 1 and 2...basically, from 20% chance of getting the right answer, to 50%. that's pretty nice, isn't it?

since B is more real, best guess - B.
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A firm is divided into four departments, each of which [#permalink]

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New post 09 Oct 2016, 10:46
mvictor wrote:
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 was never good with combinatorics questions...yet this question can be solved by applying logic...

we have 16 people, out of whom we need to select 3.
16C3 = 16*15*14/6 -> 8*5*14 => 560.
since this gives the whole # of possibilities, in which even members of the same group are counted...it must be true that the total number must be less than 560...
1. 4^3 = 64 - too little
2. 4^4 = 256 - more real
3. 4^5 = 1024 - way too big
4. 4^4 = 256. multiply by 6 -> way too big.
5. 3^6 -> 3^4 = 81. 81*9 -> 729 -> way too big, and this is not even multiplying by 4.

we are left with 1 and 2...basically, from 20% chance of getting the right answer, to 50%. that's pretty nice, isn't it?

since B is more real, best guess - B.



Dear mvictor

No issues with the answer , but do follow Bunuel 's approach here for solving such questions within minimum time -

a-firm-is-divided-into-four-departments-each-of-which-106201.html#p832274

Try to minimize calculations during the exam to as little as possible , to save time ....


PS : JUST A PIECE OF SUGGESTION , PLEASE DON'T GET ME WRONG
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A firm is divided into four departments, each of which   [#permalink] 09 Oct 2016, 10:46
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