Combinatorics help: stars and bars and beyond

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Homework Help Overview

The discussion revolves around combinatorial problems involving the distribution of distinguishable toys and identical candy bars among 12 children. Participants are exploring various scenarios regarding the number of items each child can receive and the implications of constraints on the distributions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss different methods for solving parts of the problem, including the use of cases for part c and breaking down the distribution into stages for part d. Some express confidence in their answers for parts a and b, while others question assumptions made about the distribution constraints.

Discussion Status

There is an ongoing exploration of the problem with various interpretations being considered. Some participants have provided hints and guidance on how to approach certain parts, while others are questioning the assumptions underlying the problem setup. No consensus has been reached on the correct interpretations or methods for all parts.

Contextual Notes

Participants note that part c specifies "at most 1" for candy bars, leading to discussions about whether all items must be distributed and how that affects the answers to previous parts. There is also mention of the total number of items and the implications for distribution.

Regina Fillange
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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.
 
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Regina Fillange said:
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
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.
 
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?
 
Joffan said:
d) I'm assuming that every child gets a toy.
Why do you assume that? It says at most 1.
 
Joffan said:
d) I'm assuming that every child gets a toy.

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

Well, OK. You could answer it that way too... but in c) we both assumed we kept on giving until we hit the constraints.
 
Joffan said:
Well, OK. You could answer it that way too... but in c) we both assumed we kept on giving until we hit the constraints.
Yes, and with reason:
Regina Fillange said:
c) ...distribute the ... candy bars

Regina Fillange said:
d) ... distribute ... toys
Note the definite article in c).
 
Last edited:
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? :-)
 
Joffan said:
Hmm, well firstly, it still says "at most 1" in (c),
So? k < 12 here, so it is always possible to distribute them all.
Joffan said:
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?
Excellent point. I believe you are right - Regina's answers to a) and b) are wrong.
Joffan said:
the opening sentence starts "We have to distribute m distinguishable toys, k identical candy bars to 12 children..."
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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