3 groups formed out of 6 students = 6C3 = 20vaivish 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
OA is e.
3 topics , 3 groupsrosh26 wrote:Can you explain this line:
3 topics assigned to 3 different groups = 3*3*3 = 27
wouldnt it be just 3*3?
It actually says that the 3 groups were assigned to 3 different topics, so 3*2*1 is the right calculation.parallel_chase wrote:3 topics , 3 groupsrosh26 wrote:Can you explain this line:
3 topics assigned to 3 different groups = 3*3*3 = 27
wouldnt it be just 3*3?
1st group - 3 topics
2nd group - 3 topics
3rd group - 3 topics
3*3*3 = 27
If the question would have said no two group can get the same topic then it would have been 3*2*1 = 3!, since the question doesnt mention any such condition it would be 3^3.
Hope this helps, let me know if you still have any doubts.
Stuart thanks for the clarification, i was actually wandering at my own solution for quiet sometime.Stuart Kovinsky wrote: It actually says that the 3 groups were assigned to 3 different topics, so 3*2*1 is the right calculation.
If we were selecting 3 different people out of 6 people, then 6C3 would be the correct calculation to use, but that's not what we're doing in this case - we're dividing 6 people up into 3 groups of 2. Mayur00's post explains how to do the calculations for this question.
Sure!parallel_chase wrote:
Stuart thanks for the clarification, i was actually wandering at my own solution for quiet sometime.
I am clear on the point that it should be 3!
But the way Mayur00 has calculated the groups i am quiet confused, it be nice if you could elucidate on that.
Thanks
Thanks Stuart that is very well explained. I feel after coming so far in GMAT quant that was a silly mistake to make.Stuart Kovinsky wrote:
Sure!
When we choose the first group, we have 6 people available and want to select 2 of them: 6C2.
When we choose the second group, we've already set a group of 2 aside, so now we have 4 people available and want to choose 2 of them: 4C2.
When we choose our final group, there are only 2 people remaining and we need to choose both of them: 2C2.
As with all multi-part combination questions, we MULTIPLY the individual results: 6C2 * 4C2 * 2C2.
It may be tempting to do it this way:
first group: 6 people for slot 1, 5 people for slot 2: 6*5
second group: 4 people for slot 1, 3 people for slot 2: 4*3
third group: 2 people for slot 1, 1 person for slot 2: 2*1
Total possibilities: 6*5*4*3*2*1
The problem with this approach is that it also takes into account the order in which we choose the groups, so we end up counting single combinations multiple times.
For example, if we call the 6 people A, B, C, D, E and F, using the 6! method we'd have counted:
(AB)(CD)(EF), (BA)(CD)(EF), (DC)(FE)(AB) and many other arrangements as disinct, when in fact they all end up giving us the same 3 groups.
Stuart Kovinsky wrote:If taking into the account the order in which we choose the groups does NOT matter, than the answer is 90, and if order DOES matter, then it is 540. The question comes down to whether order matters.parallel_chase wrote:
Total possibilities: 6*5*4*3*2*1
The problem with this approach is that it also takes into account the order in which we choose the groups, so we end up counting single combinations multiple times.
For example, if we call the 6 people A, B, C, D, E and F, using the 6! method we'd have counted:
(AB)(CD)(EF), (BA)(CD)(EF), (DC)(FE)(AB) and many other arrangements as disinct, when in fact they all end up giving us the same 3 groups.
You are right. But I guess order does matter here, though they should have explicitly stated it.Ian Stewart wrote:I'm curious about the source of the question, because the OA is not correct, and none of the above solutions is correct either, unfortunately. It's a bit easier to see the correct solution if you begin by imagining three topics- say Birds, Lizards and Fish. We can choose a group for Birds in 6C2 ways, then a group for Lizards in 4C2 ways, and finally a group for Fish in 1 way, using the remaining two students (of course, that's the same as 2C2). We thus have
6C2 * 4C2 * 1 = 15*6 = 90 choices in total.
If anyone is unconvinced, try doing a simpler problem: say you have a class of two, and you will divide the class into two groups of one, then assign each group a topic. Clearly there are only two ways to do this- either A has the first topic and B has the second, or A has the second topic and B has the first. There are not 4 ways to do this- if we count the ways to order A and B *and* assign them topics, we're assuming order matters not once but twice, which is a mistake.
The above solutions did not begin by selecting the topics, which is fine, but they miscounted the number of ways to divide the class into groups. When you calculate 6C2 * 4C2 * 2C2, this is only correct if the order of the groups themselves matters- that is, it is only correct if there is a 'first group', a 'second group' and a 'third group'. It would be the right answer if you were going to choose a group to work in the library, a group to work in the cafeteria and another to work in the playground- then choosing the group AB first, to work in the library, is different from choosing the group AB last, to work in the playground. But here, the order of the groups should not matter: if we choose the groups AB, CD and EF, that's the same as choosing the groups EF, CD and AB- each person is on the same team no matter what order we list the teams in. So if we count the number of possible groups in this way, we must divide 6C2 * 4C2 * 2C2 by the number of ways we can put three things in order- that is, we must divide by 3!. Then of course we simply multiply by 3! all over again, because we assign topics to the groups, which is equivalent to assigning the groups an order.
The answer should be 90, not 540.
Well, there are two possibilities; either the question-writer intended the order of the groups to matter even *before* the topics are selected and failed to communicate that clearly, or the question-writer didn't think through the question correctly. It's why I asked about the source of the question. As this question is written, if the correct answer is really '540', the question is more a test of our psychic abilities (can we guess what the question writer is thinking?) than of our mathematical abilities. Real GMAT questions are never ambiguous in this way, fortunately!canuckclint wrote: You are right. But I guess order does matter here, though they should have explicitly stated it.
First we need to select the groups:vaivish 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
