Combinatorics help: stars and bars and beyond

In summary: 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.
  • #1
Regina Fillange
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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.
 
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  • #2
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.
 
  • #3
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
Joffan said:
d) I'm assuming that every child gets a toy.
Why do you assume that? It says at most 1.
 
  • #5
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.
 
  • #6
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:
  • #7
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
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.
 

Related to Combinatorics help: stars and bars and beyond

1. What is the "stars and bars" method in combinatorics?

The "stars and bars" method is a counting technique used to solve problems involving distributing identical objects into distinct groups. It involves representing the objects as stars and using bars to represent the boundaries between groups.

2. How do I use the "stars and bars" method to solve a problem?

To use the "stars and bars" method, first determine the number of objects to be distributed and the number of groups to distribute them into. Then, draw the appropriate number of stars and bars, making sure that the number of stars equals the total number of objects. Finally, count the number of ways the stars and bars can be arranged, keeping in mind that the bars cannot be placed next to each other.

3. What is the difference between the "stars and bars" method and the "combination" method?

The "stars and bars" method is used for problems involving distributing identical objects into distinct groups, while the "combination" method is used for problems involving selecting a certain number of objects from a larger group without replacement. In the "stars and bars" method, the order of the objects within each group does not matter, while in the "combination" method, the order does matter.

4. Are there any limitations to the "stars and bars" method?

Yes, the "stars and bars" method can only be used for problems involving distributing identical objects into distinct groups. It cannot be used for problems involving objects with different characteristics or restrictions on their placement.

5. Can the "stars and bars" method be extended to solve more complex problems?

Yes, the "stars and bars" method can be extended to solve more complex problems through the use of "extended stars and bars" or other variations. These variations may involve adding additional symbols or restrictions to better represent the problem at hand.

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