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

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
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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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We are dividing by 3! because the order of groups is not important

Take a simple example of 4 people A,B,C,D being divided into 2 equal groups
The groups may be:
(A,B) and (C,D)
(A,C) and (B,D)
(A,D) and (B,C)

When we are computing the possibilities using 4C2*2c2 we are also including
(C,D) and (AB)
(B,D) and (A,C)
(B,C) and (A,D)

But we know the latter need not be counted because (A,B) and (C,D) is the same as (C,D) and (A,B)

To get the required number of possibilities without repeating, we divide by (no.of groups)!
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Ways of selecting group = 6C2 * 4C2 * 2C2 / 3! = 15

Three groups can select three subjects in 6 ways

Therefore total combinations = 15*6 = 90

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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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
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could you explain the highlighted step... i'm getting 90 = 15 * 3!

suppose the students are numbered 1,2,3,4,5,6 and tasks are X,Y and Z

one of the 15 possible ways of forming teams is 12, 34, 56. these teams can be assigned to 3 tasks in 3! = 6 ways
X-- Y-- Z
12-- 34-- 56
12-- 56-- 34
34-- 12-- 56
34-- 56-- 12
56-- 12-- 34
56-- 34-- 12
so the answer should be 15*6 = 90

i'd disagree, i dont think order of task is important here .... how do we know that 3 teams cant perform 3 tasks simultaneously....

if i "have to" arrive at 540... i'll use following logic ..

ways of arranging of 6 students in 3 teams = 6!/2!*2!*2! ( arranging 6 things in row, with 2 each in 3 groups)
= 90

each team can be assigned taskes in 3! ways

total arrangements = 90 * 3! = 540.

but i still dont agree with this answer can someone please explain..
whats the source and OE.

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

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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Number of ways of forming 3 teams = 6C2*4C2*3C2/3! = 15
Number of ways of assigning the 3 teams to 3 tasks = 3!
Number of ways of performing the 3 tasks = 3!

So total arrangements = 15*3!*3! = 540

90 is the number of ways you can assign 3 teams formed out of 12 people to 3 different tasks.
But now you can order the 3 tasks in 3! ways. T1 T2 T3 or T2 T1 T3.... etc etc.

I was confused between 90 and 540 but since question used the word "arrangements" decided to go with complete arrangements including the order of tasks.

You also need to consider if tasks assigned are X-12, Y-34, Z-56 then these tasks can be performed in following order

X-12 Y-34 Z-56
X-12 Y-56 Z-34
Y-34 X-12 Z-56
Y-34 Z-56 X-12
Z-56 Y-34 X-12
Z-56 X-12 Y-34

Therefore total combinations = 15*6*6

Yes, the answer is E

\(C_6^2 * C_4^2 * P_3\)

Which is a fair argument. But since both the options were given, and the question asked "arrangements were possible", i chose 540.

s1, (s2,s3,s4,s5,s6) - G1
s2, (s3,s4,s5,s6) - G2
s3, (s4,s5,s6) - G3
s4, (s5,s6) - G4
s5, (s6) - G5

Total groups 15.

3 topics A,B,C can be distributed in 6 (3!) different ways to 15 different groups. Hence 15*6=90, arrangements are possible.

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

The question is really tough.. I got 540 as answer but after reading the solution understood why the no. of different arrangements should be 90 only..

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?