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# Katie has 9 employees that she must assign to 3 different projects. If

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Manager
Joined: 15 Jul 2004
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Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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09 Apr 2005, 01:18
2
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Difficulty:

65% (hard)

Question Stats:

62% (02:03) correct 38% (02:15) wrong based on 440 sessions

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Katie has 9 employees that she must assign to 3 different projects. If 3 employees are assigned to each project and no one is assigned to multiple projects, how many different combinations of project assignments are possible?

A. 252
B. 1,680
C. 2,340
D. 362,880
E. 592,704
VP
Joined: 30 Sep 2004
Posts: 1149
Location: Germany
Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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09 Apr 2005, 02:26
ok take 3 out of 9 => 9c3 then 3 out of remaining 6 => 6c3 and then 3 out of remaining 3 => 3c3 => 9c3*6c3*3c3=1680
Manager
Joined: 07 Mar 2005
Posts: 143
Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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15 Apr 2005, 17:21
i get yr points guys,,,but since there are three differet projcts also to choose from so i guess it should be

(9c3*3c1)*(6c3*2c1)*(3c3*1c1)
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Joined: 15 Mar 2005
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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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17 Apr 2005, 00:44
karun_aggarwal wrote:
i get yr points guys,,,but since there are three differet projcts also to choose from so i guess it should be

(9c3*3c1)*(6c3*2c1)*(3c3*1c1)

Your point does make sense Karun.

C(9,3) makes the selection of first group of 3 from a set of 9, and then C(6,3) makes a selection of the second group of 3 from the set of remaining 6 and then C(3,3) makes a selection of the final group of 3 from the remaining 3 (trivial case obviously).

However these three groups can be allocated to the three projects in 3! ways.

I think we should have a 3! multiplier to the answer.

What say guys? Any ideas on this one?
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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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24 Nov 2013, 22:00
1
The answer should be $$9C_3*6C_3*3C_3$$

There is no need for a further factor of 3! since the order has already been taken into account. In fact, if the question had also stated that the three projects are indistinguishable, then it would have been required to divide the expression by 3! to account for having already taken the order into account.
Intern
Joined: 26 Dec 2013
Posts: 6
Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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05 Feb 2014, 12:59
1
Alternative approach:

9!/(3!*3!*3!) = 1680

We know there are a total of 9! possibilities without any groupings/restrictions.

Applying groupings/restrictions, there are 3 groups that must be made of 3 different employees (3!*3!*3!)
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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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11 Mar 2014, 05:35
1
1
The formula is nCr * (n-r)Cr * (n-2r)Cr * .... (n- (p-1)r) C r
where n is the total number of persons, r is the number of persons to be selected for each one of p groups

= 9C3 *6C3*3C3 = 1680

Explanation: Initially there are 9 persons. You can select any 3 persons to assign to project 1 or project 2 or project 3. Assume you assign them to project 1. Now there are 6 persons left and you can assign any 3 of them to project 2 or project 3. Assume you assign them to project 2. Finally there are 3 persons left with all those 3 persons to be assigned to project 3.

In the above you chose project 1 first, project 2 next and project 3 last. This project selections can be done in3! ways. Suppose had you assigned them to project 2 first then to project 1 and then to project 3 the same set of arrangements would happen. So you do not have to multiply by 3!. Generally speaking when you take care that any group of r persons may be assigned for each slot/task in arriving at the answer then this above ordering is not required.
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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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15 Mar 2014, 19:45
EE Combos for (Project 1 * Project 2 * Project 3) / project permutations (corrective measure to eliminate over-counting)
3C9*3C6*3C3/3P2
=1680
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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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04 Jul 2015, 11:00
Zem wrote:
Katie has 9 employees that she must assign to 3 different projects. If 3 employees are assigned to each project and no one is assigned to multiple projects, how many different combinations of project assignments are possible?

A. 252
B. 1,680
C. 2,340
D. 362,880
E. 592,704

The ways to select three employees out of 9 for first project = 9C3

The ways to select three employees out of Remaining 6 for Second project = 6C3

Third Project already has the group of three to be assigned to it so need to select that

So total ways to assign 3 employees to each of 3 projects = 9C3 * 6C3 * 1 = 1,680

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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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03 Jan 2016, 19:45
$$9C_3*6C_3*3C_3*3 projectsC_3groups$$

Bunuel or some Quant guru, is the logic of the final term correct (Assigning 3 projects to 3 groups).
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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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03 Jan 2016, 21:47
TooLong150 wrote:
$$9C_3*6C_3*3C_3*3 projectsC_3groups$$

Bunuel or some Quant guru, is the logic of the final term correct (Assigning 3 projects to 3 groups).

What you have done is CORRECT because 3C3 is 1 anyways
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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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25 Aug 2017, 11:13
Why do we not divide by 3! when in this problem: https://gmatclub.com/forum/in-how-many- ... 01722.html

we divide by 3! ?
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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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11 Aug 2019, 19:24
Zem wrote:
Katie has 9 employees that she must assign to 3 different projects. If 3 employees are assigned to each project and no one is assigned to multiple projects, how many different combinations of project assignments are possible?

A. 252
B. 1,680
C. 2,340
D. 362,880
E. 592,704

The first 3 employees can be selected in 9C3 = (9 x 8 x 7)/3! = 3 x 4 x 7 = 84 ways.

The next 3 employees can be selected in 6C3 = (6 x 5 x 4)/3! = 20 ways.

The final 3 employees can be selected in 3C3 = 1 way.

Therefore, the total number of ways the 3 projects can be assigned is:

84 x 20 x 1 = 1,680

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Re: Katie has 9 employees that she must assign to 3 different projects. If  [#permalink]

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11 Aug 2019, 19:24
Zem wrote:
Katie has 9 employees that she must assign to 3 different projects. If 3 employees are assigned to each project and no one is assigned to multiple projects, how many different combinations of project assignments are possible?

A. 252
B. 1,680
C. 2,340
D. 362,880
E. 592,704

The first 3 employees can be selected in 9C3 = (9 x 8 x 7)/3! = 3 x 4 x 7 = 84 ways.

The next 3 employees can be selected in 6C3 = (6 x 5 x 4)/3! = 20 ways.

The final 3 employees can be selected in 3C3 = 1 way.

Therefore, the total number of ways the 3 projects can be assigned is:

84 x 20 x 1 = 1,680

_________________

# Scott Woodbury-Stewart

Founder and CEO

Scott@TargetTestPrep.com
122 Reviews

5-star rated online GMAT quant
self study course

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

If you find one of my posts helpful, please take a moment to click on the "Kudos" button.

Re: Katie has 9 employees that she must assign to 3 different projects. If   [#permalink] 11 Aug 2019, 19:24
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