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I'm having trouble with the following.

  1. A man has 10 distinct candies and puts them into two distinct bags such that each bag contains 5 candies. In how many ways can he do it?

a. For this problem I thought it would (10 choose 5) since we could place 5 candies in one box out of 10, and then for the second box 5 candies would be left, so (5 choose 5) however im not sure if it is right.

  1. How many ways are there to divide 10 boys into two basketball teams of 5 boys each?

b. I would think this is similar to A, and am not sure.

  1. A person has 10 distinct candies, and puts them in two identical bags, such that no bag is empty, how many ways can he do it?

  2. A person has 10 identical candies and puts them in two identical bags such that none are empty, how many different ways can he do it.

I'm practicing for a exam and can't figure them out.

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candies $1,2,3,4,5$ in bag $1$ and $6,7,8,9,10$ in bag $2$, is that here the same as candies $1,2,3,4,5$ in bag $2$ and $6,7,8,9,10$ in bag $1$? If the bags are explicitly distinct then the answer is 'no' and your thinking is okay. What if you ask the same sort of question concerning the boys? –  drhab Nov 19 '13 at 16:53
    
removing the revision tag per discussion here: meta.math.stackexchange.com/questions/11694/… –  Willie Wong Nov 22 '13 at 11:16

3 Answers 3

With a and b, it depends whether the bags/teams are distinct. Probably the bags are, and your answer to a is correct. For b, the teams are not distinct, so you need to divide by two, because picking one group of five gives the same teams as picking the other group of five.

For 1, you can pick any set of the candies to put in the first bag except all or none of them. How many subsets is that? Then divide by 2 because of swapping the bags. For 2, all that matters is how many candies are in each bag. How many ways can you add two numbers to make ten? Again you need to account for the bags being identical.

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Hint concerning 10 distinct candies in identical bags:

First count how many ways there are if the bags are distinct (your thoughts about that are correct). Then realize that - if there is no distinction between the bags - every way is counted twice.

Hint concerning $10$ identical candies in identical bags (not empty):

$10=1+9=2+8=...$

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I’m going to add a little more explanation of the difference between the first and second problems. In the first problem we’re told that the bags are distinct; given the wording of the last two problems, that almost certainly means that they are individually identifiable, not interchangeable. That’s as if in the second problem we were to split the $10$ boys into a team called the Gryphons and another team called the Hippogryphs. There would be $\binom{10}5$ ways to choose which $5$ boys are to be Gryphons, and of course the other $5$ boys would be the Hippogryphs, so there would be $\binom{10}5$ ways to choose the teams. Similarly, in the first problem there are $\binom{10}5$ ways to split the candies between the two individually identifiable bags.

In the actual second problem, however, the teams don’t have names; they’re just two teams of $5$. If the boys are $A,B,C,D,E,F,G,H,I$, and $J$, and we pick $A,B,C,D$, and $E$ to be one team, then of course $F,G,H,I$, and $J$ make up the other team. But we might just as well have selected $F,G,H,I$, and $J$ for one team, leaving $A,B,C,D$, and $E$ to be the other team. Since the teams are not named, choosing $A,B,C,D$, and $E$ gives us exactly the same division into teams as selecting $F,G,H,I$, and $J$: every possible division into teams is counted twice. That’s why in the second problem you don’t get $\binom{10}5$ possibilities, but rather only $\frac12\binom{10}5$. To put it a little differently, picking $A,B,C,D$, and $E$ to be the Gryphons in the modified problem gives us a different division into named teams from picking $F,G,H,I$, and $J$ to be the Gryphons, but it gives us exactly the same two groups of people playing against each other. It’s a different division into named (or otherwise individually identifiable) teams but not a different division into two teams of $5$.

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