Probability Question gmat prep 2

[accounts]

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

[gokul_nair1984]

There will be 3 groups with 2 student in each group.
Total number of Students=6

Any 2 students can be selected in C(6,2) ways ie;15 ways.
Any one group out of 3 can be chosen in 3! ways.

Therefore total number of ways of forming 3 groups with 2 students each=15*3!=90

Furthermore, One course topic can be selected from 3 topics in 3! ways.

Thus the number of ways the three groups could be assigned to three different topics=90*3!=540

[accounts]

Gokul
But should we take into account these 3 groups of 2 ...

my answer was

(6c2 x 4c2 x 2c2 / 3! ) * 3! =90

we need to divide by 3! because there are 3 groups with same number of people.

the assigning of subjects 3! ways is correct.

[gokul_nair1984]

my answer was

(6c2 x 4c2 x 2c2 / 3! ) * 3! =90

we need to divide by 3! because there are 3 groups with same number of people.

The number of people might be the same but they are distinct from one another( unlike a group of red balls or a group of white balls).Hence you need not divide by 3!

Let me put this in another way.

1.To form the first group, we need to select 2 people from 6. This can be done by C(6,2) number of ways.
2.To form the second group, we need to select 2 people form the remaining 4(6-2). This can be done by C(4,2) number of ways.
3.Finally, to form the last group, we have to select 2 people from the remaining 2(4-2).This can be done in only 1 way ie;C(2,2)

Multiply this entire thing by 3! and you get your answer.I hope you could understand.

[RonPurewal]

Gokul
But should we take into account these 3 groups of 2 ...

my answer was

(6c2 x 4c2 x 2c2 / 3! ) * 3! =90

this is correct.

notice that, as is the case for a great many combinatorics problems on this test, you can also solve this problem by just making a list and counting the items in the list.
if you are not sure how to break the six students down into three groups, just start listing different breakdowns. let's say that the students are numbered 1, 2, 3, 4, 5, 6.
then the possible breakdowns are
12 34 56
12 35 46
12 36 45
13 24 56
13 25 46
13 26 45
14 23 56
14 25 36
14 26 35
15 23 46
15 24 36
15 26 34
16 23 45
16 24 35
16 25 34
... and that's all of them. 15 of them.
then you have to assign the three topics to these three groups, so that entails multiplication by 3!, giving 15 x 3! = 90.
so the correct answer is 90.

--

note that there is actually a much more straightforward solution to this problem.
instead of worrying about selecting the three groups of students AND THEN assigning them to the three topics, you can just assign the groups to the topics right away! notice that this does away with the issue of order -- thereby making the problem much simpler -- since the groups are now distinguishable from each other.
first, select the group that works on topic #1. this can be done in 6c2 = 15 ways.
then, select the group that works on topic #2. this can be done in 4c2 = 6 ways, since there are four people left over from whom to choose.
at this point, you're done, since there are only two people left for the last group. if you're a purist, though, you can always write out 2c2 for the last group (although this is a complete waste of time).
so, 15 x 6 = 90.

--

ans = 90

[gokul_nair1984]

I realised my error.

instead of worrying about selecting the three groups of students AND THEN assigning them to the three topics, you can just assign the groups to the topics right away! notice that this does away with the issue of order -- thereby making the problem much simpler -- since the groups are now distinguishable from each other.
first, select the group that works on topic #1. this can be done in 6c2 = 15 ways.
then, select the group that works on topic #2. this can be done in 4c2 = 6 ways, since there are four people left over from whom to choose.
at this point, you're done, since there are only two people left for the last group. if you're a purist, though, you can always write out 2c2 for the last group (although this is a complete waste of time).
so, 15 x 6 = 90.

This is much simpler to comprehend
Thank You Ron

[RonPurewal]

I realised my error.

instead of worrying about selecting the three groups of students AND THEN assigning them to the three topics, you can just assign the groups to the topics right away! notice that this does away with the issue of order -- thereby making the problem much simpler -- since the groups are now distinguishable from each other.
first, select the group that works on topic #1. this can be done in 6c2 = 15 ways.
then, select the group that works on topic #2. this can be done in 4c2 = 6 ways, since there are four people left over from whom to choose.
at this point, you're done, since there are only two people left for the last group. if you're a purist, though, you can always write out 2c2 for the last group (although this is a complete waste of time).
so, 15 x 6 = 90.

This is much simpler to comprehend
Thank You Ron

glad it helped

[agha79]

Gokul
But should we take into account these 3 groups of 2 ...

my answer was

(6c2 x 4c2 x 2c2 / 3! ) * 3! =90

we need to divide by 3! because there are 3 groups with same number of people.

the assigning of subjects 3! ways is correct.

how is this (6c2 x 4c2 x 2c2 / 3! ) * 3! =90 is made?
number of groups is 6c2 which is 20. 3 subjects that is 3! so my understanding was 20*6=120.

[agha79]

Ooops sorry my mistake 6c2 is 15.
The number of groups=3. Possible ways of group is 3! So instead of the long formula used in one of the previous posts cant we just multiple 15 with 3! That gives 90?

[jnelson0612]

Ooops sorry my mistake 6c2 is 15.
The number of groups=3. Possible ways of group is 3! So instead of the long formula used in one of the previous posts cant we just multiple 15 with 3! That gives 90?

Correct, 15 * 3! is 15 * 3 * 2 * 1 which is 90.