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how can I solve this:

How many different ways can 8 children be divided into two groups of 3 and one group of 2?

My method was:

8C3*5C3*2C2+8C3*5C2*3C3+8C2*6C3*3C3=560+560+560=1680

But the answer is:

8C3*5C3*2C2=560

I feel close to getting the logic but I always seem to miss it... Can anyone point it out to me? I'm bad in perms and combs. Thanks.

user9856
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2 Answers2

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It doesn't matter what order you pick out the groups. So first picking $3$ children, then $3$ children then $2$ (the first term in your sum) yields a result that in the end is indistinguishable from any of the two other orders you can do it.

Which group you pick out children for first is irrelevant to what groups you end up with at the end, and as such having one term for each way it can be done leads to overcounting.

Arthur
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  • So does the "doesn't matter what order you pick out the groups" apply to every question using C? (combinations) – user9856 Sep 28 '14 at 10:48
  • No, it has nothing to do with the C (that is, the internal ordering within a group and whether it matters), but rather the fact that the end result doesn't care which groups are declared in what order. – Arthur Sep 28 '14 at 10:53
  • Oh so I guess this is why the textbooks always say "Order doesn't matter" in Combinations...Thanks! – user9856 Sep 28 '14 at 10:56
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    Yes. To be fair, deciding when order does and does not matter, and how it impacts the calculations can be quite difficult at times, and it's definitely the part of combinatorics I've had most trouble with when teaching – Arthur Sep 28 '14 at 11:01
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The logic behind here is the set of actions that take place.

First, we have $8$ people choosing $3$ kids for the first group. That is $8C3$

Next, we have $5$ kids left, who need to be divided into $3$ kids to form the second group.

Finally, we only have $2$ kids left and one group $2$, so that is $2C2$, or $1$ way to arrange the kids.

So, we get that:

$$8C3*5C3*2C2 = 560$$

Varun Iyer
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