is how many ways could a class of 18 students divide into groups of 3 students each~ please leave full answers

thank u

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- Mar 25th 2008, 05:49 AM #1

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- Mar 25th 2008, 08:12 AM #2

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I will assume with the title of your thread that we don't need "unique" arrangements, we are dealing with groups of 3 choosing from a pool of 18: That answer is 4,896.

Go here:

Permutation and Combination Calculator

Number of Items is 18, Arrangements is 3. This is read as 18 choose 3.

If in fact I am incorrect in my assumption, and we need unique arrangements, press the combination button instead. That would be read as 18 choose 3 unique. That answer would be 816.

Either way will show you the math work for your answer

Think of it as having 18 people, 1-18. We take groups of 3, such as (1,2,3), (4,5,6), etc. For arrangements that do not need to be unique, (1,2,3) is considered different than (3,2,1). For unique arrangements, i.e., combinations, (1,2,3) is considered the same as (3,2,1) so that type is only counted once.

Any questions?

- Mar 25th 2008, 08:45 AM #3
Unfortunately the answer above is incorrect.

The wording of the question clearly implies that groups are not labeled. Therefore, this is known as*unordered partition*.

The answer is $\displaystyle \frac{{18!}}{{\left( {6!} \right)\left( {3!} \right)^6 }} = {\rm{190590400}}$.

If the groups are labeled such, say by team names, then it is a ordered partition and the answer is $\displaystyle \frac{{18!}}{{\left( {3!} \right)^6 }} $.

- Mar 25th 2008, 09:05 AM #4

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I'm on wikipedia reading about unordered partitions. I don't see a good reference on this. Can you provide a link for me to read up on this?

Here is where I'm at right now.

Twelvefold way - Wikipedia, the free encyclopedia

Towards the bottom where it has S(n,x)

- Mar 25th 2008, 09:48 AM #5
This is one of many cases in mathematics where terms have several uses.

What you are reading about, Stirling numbers, help in counting partitions of sets. But this problem is strictly about a partition of a set of 18 into cells of 3 each. The in general problem of partitions we consider how many ways to partition a set into nonempty, non-overlapping cells.

- Mar 25th 2008, 10:03 AM #6

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## thank you

hehe my data teacher give out the answer but he did it wrong... my god.......... and he said the right answer of the book is wrong_

luckily, i have got it myself, but i did it in the long way, i didn't use the (3!)^6

because i just learn the staff of the permutation, that question is really tough_ thank you very much still ```