Dogs do not share biscuits!! Combinatorial Problem
Ten indistinguishable dogs encounter eight indistinguishable biscuits. We all know that the dogs do not share biscuits. What is the number of different ways the biscuits can be consumed?
It is easy to see that this could happen in C(17,8) ways if we allow dogs to be distinguishable. Now dividing this answer by 10! seemed very reasonable to me. But if we set #dogs=3 and #biscuits = 2 then we get answer of one applying my logic. Clearly Wrong!!!
Could anyone help me on this one?
Thank you!!
It is easy to see that this could happen in C(17,8) ways if we allow dogs to be distinguishable. Now dividing this answer by 10! seemed very reasonable to me. But if we set #dogs=3 and #biscuits = 2 then we get answer of one applying my logic. Clearly Wrong!!!
Could anyone help me on this one?
Thank you!!
Off the top of my head, I don't think your logic works (and you checked and it doesn't), but don't have a quick/easy explanation why.
The answer should just be the number of unordered partitions of 8 as you don't care which dog eats which biscuit, so something like
{{8},
{7,1},
{6,2},
{6,1,1},
{5,3}, ...}
Should be 22 unordered partitions?
The answer should just be the number of unordered partitions of 8 as you don't care which dog eats which biscuit, so something like
{{8},
{7,1},
{6,2},
{6,1,1},
{5,3}, ...}
Should be 22 unordered partitions?
To try and see where your intuition doesn't work, consider 10 dogs and 1 biscuit.
C(10,9) = 10,
i.e. a dog has a biscuit and the remaining 9 go hungry (10 times).
Dividing by 10! is a bit too much. If you have all the dogs having biscuits all the time, that should be ok, but obviously some scenarios have unhappy, unfed dogs and these would need to be left out of the divisor to go from ordered/distinguishable to unordered/undistinguishable
.
C(10,9) = 10,
i.e. a dog has a biscuit and the remaining 9 go hungry (10 times).
Dividing by 10! is a bit too much. If you have all the dogs having biscuits all the time, that should be ok, but obviously some scenarios have unhappy, unfed dogs and these would need to be left out of the divisor to go from ordered/distinguishable to unordered/undistinguishable
.
(edited 5 years ago)
Original post by mqb2766
Off the top of my head, I don't think your logic works (and you checked and it doesn't), but don't have a quick/easy explanation why.
The answer should just be the number of unordered partitions of 8 as you don't care which dog eats which biscuit, so something like
{{8},
{7,1},
{6,2},
{6,1,1},
{5,3}, ...}
Should be 22 unordered partitions?
The answer should just be the number of unordered partitions of 8 as you don't care which dog eats which biscuit, so something like
{{8},
{7,1},
{6,2},
{6,1,1},
{5,3}, ...}
Should be 22 unordered partitions?
PRSOM and Thank you!
Sorry for not addressing the reply earlier. I haven't come across unordered partitions but primarily listing all the possibilities (tedious task) I could verify it was 22.
Original post by mqb2766
To try and see where your intuition doesn't work, consider 10 dogs and 1 biscuit.
C(10,9) = 10,
i.e. a dog has a biscuit and the remaining 9 go hungry (10 times).
Dividing by 10! is a bit too much. If you have all the dogs having biscuits all the time, that should be ok, but obviously some scenarios have unhappy, unfed dogs and these would need to be left out of the divisor to go from ordered/distinguishable to unordered/undistinguishable
.
C(10,9) = 10,
i.e. a dog has a biscuit and the remaining 9 go hungry (10 times).
Dividing by 10! is a bit too much. If you have all the dogs having biscuits all the time, that should be ok, but obviously some scenarios have unhappy, unfed dogs and these would need to be left out of the divisor to go from ordered/distinguishable to unordered/undistinguishable
.
Thanks for the explanation. Same classic error of over counting and over-dividing.
Original post by Quantum Horizon
PRSOM and Thank you!
Sorry for not addressing the reply earlier. I haven't come across unordered partitions but primarily listing all the possibilities (tedious task) I could verify it was 22.
Thanks for the explanation. Same classic error of over counting and over-dividing.
Sorry for not addressing the reply earlier. I haven't come across unordered partitions but primarily listing all the possibilities (tedious task) I could verify it was 22.
Thanks for the explanation. Same classic error of over counting and over-dividing.
There is a fibonacci-type recurrence relationship for generating the number of unordered partitions, as well as a bit more info at:
https://en.wikipedia.org/wiki/Partition_(number_theory)
Quick Reply
Related discussions
- I am a giant Union Jack wearing toy bulldog AMA!
- Human foods for dogs
- Name a …. that doesn’t contain the letter
- Assistance Dog at Uni
- issue w/ international student flatmates
- Favourite places to walk your dog
- What trips have you been on animal behaviour/zoology degree
- Bonfire Night: How to keep pets safe & happy during fireworks
- Muslim Girl but I want to be a (animal) Vet
- Vet med dissections
- I need advice please!!!!
- NEA 1 task GCSE AQA Food Preparation and Nutrion
- Guide Dogs Trainee Guide Dog Trainer
- Summer Internship Accommodation London
- University
- Out of Control Animal
- Is jaffa cake a cake or biscuit?
- Discontinued favourite foods or snacks.
- Maths Personal Statement HELPPP
- hi
Latest
Trending
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
Trending
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
