# Combinatorics help: stars and bars and beyond

1. Dec 7, 2014

### Regina Fillange

• Member warned about not using the homework template
We have to distribute m distinguishable toys, k identical candy bars to 12 children in the following ways:

a. How many ways can we distribute toys if each child can get any number of toys?
b. How many ways can we distribute candy if each child can get any number of candy bars?
c. How many ways can we distribute the (identical) candy bars if each child can get at most 1 and k<12
d. How many ways can we distribute (distinct) toys if each child can get at most 1 and m>12 (extra toys are saved for next year)
e. If m=15 and k=9 how many ways can we distribute the objects if each child gets exactly 2 goodies.

Parts a and b I feel confident about (I got 12^m for part a and C(k+11, k) for part b).

Part c I tried doing by cases, such as 11 children can have 0 or 1, then 10 children can have 0 or 1 etc. So I figured it would be 2^11

Parts d and e I truly feel lost on and have scoured the internet for different methods and haven't felt confident about anything I've read! Thank you for any starting hints or tips.

2. Dec 8, 2014

### haruspex

We have to distribute all k, yes? Given that you solved a and b, you must be overthinking c.
For d, how many toys will be distributed? Break it into two stages correspondingly.
For e, separate the distribution of toys from that of candies.

3. Dec 8, 2014

### Joffan

a) and b) look good.

c) For sure your answer should involve k somehow. Unless we're holding back some of the candy, as haruspex checks, how many children get candy?
d) I'm assuming that every child gets a toy.
e) Since m+k=24, exactly the number of gifts required, how many options are there for the candy after the toys are distributed?
Alternative e) Is there perhaps a specification that each child must get a toy as one of the gifts?

4. Dec 8, 2014

### haruspex

Why do you assume that? It says at most 1.

5. Dec 8, 2014

### Joffan

Well, OK. You could answer it that way too... but in c) we both assumed we kept on giving until we hit the constraints.

6. Dec 8, 2014

### haruspex

Yes, and with reason:
Note the definite article in c).

Last edited: Dec 8, 2014
7. Dec 8, 2014

### Joffan

Hmm, well firstly, it still says "at most 1" in (c), and secondly, does that mean Regina's answers to (a) and (b) are wrong - that as well as the children there is effectively also a box for ungifted items?

By contrast, the opening sentence starts "We have to distribute m distinguishable toys, k identical candy bars to 12 children..."

Jeez, who knew writing a problem was so hard? :-)

8. Dec 8, 2014

### haruspex

So? k < 12 here, so it is always possible to distribute them all.
Excellent point. I believe you are right - Regina's answers to a) and b) are wrong.
True, but I would take the individual instructions as more reliable. E.g. in d) it specifically states that not all need to be distributed.

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