Six students are equally divided into 3 groups, then, the three groups

Question

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

JeffTargetTestPrep · Accepted Answer

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
 
Answer: C

isaacconnor · Answer

Rephrase the question:
How many 2 person arrangements can you make out of 6 people = 6C2= 15. Then how many ways can you assign the 2 person teams to 3 assignments= 3!

Answer: 6C2 * 3! = 90

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nmohindru · Answer

Ways of selecting group = 6C2 * 4C2 * 2C2 / 3! = 15

Three groups can select three subjects in 6 ways

Therefore total combinations = 15*6 = 90

SVaidyaraman · Answer

1. Take a simple example of 4 students into 2 groups of 2 students each and two different topics being assigned .
2. Taking an example case let the students be s1,s2,s3,s4. the groups could be (s1,s2) and (s3,s4), (s1,s3) and (s2,s4) , (s1,s4) and (s2,s3), Number of ways of forming groups is 4C2*2C2/2!=3 where 2 in the denominator represents the number of groups. Remember order of groups is not important.
3. Applying the above rule to the current case, number of ways of forming groups is 6C2*4C2/3!.
4. Three topics can be arranged between the groups in 3! ways.
5. Total number of possible arrangements is (3)*(4)=90

Nitish7 · Answer

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

Nitish7 · Answer

Hi,

Thanks for your valuable feedback please let me know one more thing
as the question asks us to create 3 equal groups from 6 students so in each group there will be 2 students so if I use 6C2 will it be correct?

SVaidyaraman · Answer

It is the number of ways of first selecting 2 students out of 6 and then 2 out of the remaining 4 and then 2 out of the remaining 2. So it is 6C2*4C2*2C2. . But since order is not important, as I explained in my previous post, 6C2*4C2*2C2 has to be divided by (no.of groups)!

gmathopeful19 · Answer

There's a lot of misinformation/over complicated methods of getting the answer.

You want number of combinations of 2 people out of 6 = 6C2 = 15

15 * 3! = 90

C

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hellosanthosh2k2 · Answer

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

GMATGuruNY · Answer

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

C

rahulkhera324 · Answer

We use combination method when the order of selection doesn't matter. So if you have used combination formula then you don't require to divide by 3!. 
Please correct me if I am wrong... according to me 540 should be the answer. 
 Bunuel  can you please confirm the answer. I think it should be 540.

basicallybhavesh · Answer

Imagine you have A, B, C, D, E, F (6 students)

We used 6C2 to find the number of students in the first group, here (say) AC or CA don't matter.

However, when we have the three groups (say)

AC, BD, FE
BD, AC, FE
...
...

These are same (first two rows), hence, we need to divide it by 3!.

mbaaccount1234 · Answer

Hey Jeff! How would the question be written if order did not matter?

manan01 · Answer

Instead of solving this question wrt students, we can solve this question from the Topics View,

For Topic 1 -> we need two students from 6, 6C2 = 15.
For Topic 2 -> we need two students out of remaining, 4, 4C2 = 6
For Topic 3 -> we need two students out of remaining, 2, 2C2 = 1

Total ways of assignment = Topic 1 AND Topic 2 AND Topic 3 = 15 x 6 x 1 = 90 ways

Six students are equally divided into 3 groups, then, the three groups

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Six students are equally divided into 3 groups, then, the three groups

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615 to 715 in 15 Days (Ria's Strategies)

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