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

Three groups can select three subjects in 6 ways

Therefore total combinations = 15*6 = 90

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

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

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

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

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.

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

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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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!?

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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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?

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)!
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Imo C
3 groups can be formed in 15 ways then 3 topics can be chosen in 6 ways
So number of arrangements =90

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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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