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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Six students are equally divided into 3 groups, then, the three groups

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Target Test Prep Representative G
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2809
Re: Six students are equally divided into 3 groups, then, the three groups  [#permalink]

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arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?

(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

First we need to select the groups:

Since we have 6 students, the first group can be formed in 6C2 = (6 x 5)/2! = 15 ways. Since there are now 4 students left, the 2nd group can be selected in 4C2 = (4 x 3)/2! = 6 ways. Since there are 2 students left, the final group can be selected in 2C2 = 1 way.

Thus, the total number of ways to select the 3 groups is 15 x 6 x 1 = 90 if the order of selecting these groups matters. However, the order of the selection doesn’t matter, so we have to divide by 3! = 6. Thus, the total number of ways to select the groups when order doesn’t matter is 90/6 = 15.

Now we need to determine the number of ways to assign those 3 groups to 3 different topics, which can be done in 3! = 6 ways.

Thus, the total number of ways to form the groups and assign them to a task is:

15 x 6 = 90

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Six students are equally divided into 3 groups, then, the three groups  [#permalink]

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Total number of ways forming 3 groups of 2 each from 6 students.
= (6c2 * 4c2 * 2c2)/3!

reason we are dividing by 3! is, to eliminate this {AB, CD, EF} {AB, EF,CD}{CD,AB,EF}{CD,EF,AB}{EF,AB,CD}{EF,CD,AB}
There 6 redunduncies, to remove them we divide by 3! = 6

But the question mentions each group will different topics, permutation of topics across 3 groups is 3!

so final answer = 6c2 * 4c2 * 2c2 = 90
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Joined: 10 Apr 2018
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Re: Six students are equally divided into 3 groups, then, the three groups  [#permalink]

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Nitish7 wrote:
SravnaTestPrep
Why are we dividing the 6c2* 4c2 * 2c2 by 3!?

Well,
Just want to add my two cents as to how it has been solved on forum and why has that way been used.

lets mark the students numerically as (1,2,3,4,5,6)
How can we select the groups
Ask you self say if i pick (1,2) will the group be different if i pick (2,1) . No. So order does not matter here .
Now this is how i can pick two students for my first group
(1,2),(1,3),(1,4),(1,5),(1,6), (2,3),(2,4),(2,5),(2,6),(3,4)(3,5)(3,6)(4,5)(4,6)(5,6) So total 15 Ways

Say i pick (1,2)

Now how many are left ( 3,4,5,6). Now how can i pick two students for my second group
(3,4)(3,5)(3,6)(4,5)(4,6)(5,6) So total 6 ways

say i this time pick (3,4)

Now how many are left ( 5,6). Now how can i pick two students for my last group
only one way right (5,6) so i ways

So I pick first group AND second group AND third group

so it will be 15 X 6 X 1 =90 ways .
Why are we dividing by 3! when we use formula:

Does the order of group selection matter . Will it be any different if i selected (3,4) for my first group and then (1,2) for second group and (5,6) the last.
No.
But wait , haven't we considered (3,4)(3,5)(3,6)(4,5)(4,6)(5,6) twice . Once in first group and once in second group. So there are 6 cases which have been considered twice. Now if i had picked (5,6 ) initially i would still have 6 cases which would have been considered twice. [(1,2),(1,3),(1,4), (2,3),(2,4),(3,4)}]To get rid of duplication cases we divide by 6.

( hint in the first step we have already identified the cases in which can form groups)

So we have division by 3! in denominator.

Now lets name the group $$G_1, G_2, G_3$$

In how many ways can i assign the projects
Does the order matter . I mean say the order in which project is assigned is $$(P_1,P_2,P_3)$$ to ($$G_1,G_2,G_3$$) be any different from saying the order of projects is $$(P_1,P_2,P_3)$$ to ($$G_2,G_1,G_3$$). Yes the order matters.

SAY Project 1 can be assigned to any of the three groups . So we have 3 ways
Now project 2 can be assigned to any of the remaining two groups after one of the three group has been awarded project 1 . So we have 2 Ways
Project 3 can be assigned only one way

So total 6 ways we can assign project to three groups.

So total number of ways we can do the required is 15 Ways of distributing people in 3 groups X 6 ways of distributing 3 projects among the three formed groups.

= 90

Hope this helps

Probus
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Re: Six students are equally divided into 3 groups, then, the three groups  [#permalink]

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Suppose the six students are 1,2,3,4,5,6

Now student 1 HAS to be paired with someone else. He/she can be paired with 2,3,4,5, or 6. So that's 5 possibilities.

Once that has occurred (without loss of generality suppose 1 pairs with 2. 3 can then pair with 4,5, or 6. That's 3 possibilities. Only one pair is left. 5x3=15 groups of 3.

For each group there are 3x2x1 different arrangements of the topics. 15x6=90.
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Re: Six students are equally divided into 3 groups, then, the three groups  [#permalink]

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arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?

(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

Can someone please explain to me why the red part of 15*3!, because I really can't figure out the reasoning behind the permutation of the assignments,
guaranteed kudos for accurate responses

Thanks
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Joined: 04 Aug 2010
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Re: Six students are equally divided into 3 groups, then, the three groups  [#permalink]

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arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?

(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

2 students must be assigned to each topic, with no student assigned to more than 1 topic.
From 6 students, the number of ways to choose 2 for the first topic = 6C2 = (6*5)/(2*1) = 15.
From the 4 remaining students, the number of ways to choose 2 for the second topic = 4C2 = (4*3)/(2*1) = 6.
The 2 remaining students must be assigned to the third topic.
To combine the options in blue, we multiply:
15*6 = 90

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Six students are equally divided into 3 groups, then, the three groups  [#permalink]

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arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?

(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

6C2*4C2*2C2/3!*3! = 15*6 = 90

IMO C
Director  V
Joined: 24 Oct 2016
Posts: 586
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38 Re: Six students are equally divided into 3 groups, then, the three groups  [#permalink]

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arjtryarjtry wrote:
Six students are equally divided into 3 groups, then, the three groups were assigned to three different topics. How many different arrangements are possible?

(A) 30
(B) 60
(C) 90
(D) 180
(E) 540

Method: Anagram Grid

AA BB CC

6!/(2!2!2!) = 90 Re: Six students are equally divided into 3 groups, then, the three groups   [#permalink] 24 Nov 2019, 18:29

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