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# A firm is divided into four departments, each of which cont

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

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21 Sep 2009, 09:21
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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
[Reveal] Spoiler: OA

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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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21 Sep 2009, 10:08
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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 ...

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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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21 Sep 2009, 12:30
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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

[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.
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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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21 Sep 2009, 20:24
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.
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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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07 Jul 2010, 20:16
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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

Cheers
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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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12 Jul 2010, 16:56
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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: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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13 Jul 2010, 05:30
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.

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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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13 Jul 2010, 07:59
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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.

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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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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14 Jul 2010, 21:03
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?
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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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31 Aug 2011, 07:02
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
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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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31 Aug 2011, 09:34
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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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31 Aug 2011, 09:52
Thanks a lot for explaining the reverse combination approach.It helped me to understand combinations better.
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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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29 Sep 2011, 00:58
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?
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Re: Combinations Problem - Extreme Challenge - A firm is ... [#permalink]

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30 Sep 2011, 01:33
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.
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Re: A firm is divided into four departments, each of which cont [#permalink]

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12 Nov 2013, 06:23
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Re: A firm is divided into four departments, each of which cont [#permalink]

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12 Nov 2013, 08:12
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
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Re: A firm is divided into four departments, each of which cont [#permalink]

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30 Dec 2013, 08:00
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
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