(a) We have three standard 6-sided dice, colored red, yellow, and green. In how many ways can we roll them to get a sum of 9? The dice are colored so that a red 2, yellow 3, and green 4 is a different roll from red 3, yellow 4, and green 2.

(b) We have 10 standard 6-sided dice, all different colors. In how many ways can we roll them to get a sum of 20?

Jz1234 Aug 25, 2017

#1**+3 **

a) We have three standard 6-sided dice, colored red, yellow, and green. In how many ways can we roll them to get a sum of 9? The dice are colored so that a red 2, yellow 3, and green 4 is a different roll from red 3, yellow 4, and green 2.

These numbers are small enough so I can just count them

Green | number left | Y/R | Y/R
| Number of combination |

1 | 8 | 2 | 6 | 2 |

1 | 3 | 5 | 2 | |

1 | 4 | 4 | 1 | |

2 | 7 | 1 | 6 | 2 |

2 | 5 | 2 | ||

3 | 4 | 2 | ||

3 | 6 | 1 | 5 | 2 |

2 | 4 | 2 | ||

3 | 3 | 1 | ||

4 | 5 | 1 | 4 | 2 |

2 | 3 | 2 | ||

5 | 4 | 1 | 3 | 2 |

2 | 2 | 1 | ||

6 | 3 | 1 | 2 | 2 |

TOTAL | 25 |

(b) We have 10 standard 6-sided dice, all different colors. In how many ways can we roll them to get a sum of 20?

Yea you can think about this one by thinking of the logic that happened with my first one.

I might think on it later. :)

Melody Aug 25, 2017

#2**+2 **

(b) We have 10 standard 6-sided dice, all different colors. In how many ways can we roll them to get a sum of 20?

Ive been off counting (yes I do I agree, it is not a good use of my time )

This is what I have got

I rearranged this and got:

I am reasonably confident that my method is correct but there could be lots of careless mistakes.

Heureka often comes along and fixes it up when I do questions like this :)

Anyway I get **81,408 **different ways of getting a total of 20 with 10 unique dice :)

If you really want me to give a link to my spreadsheet I should be able to do that :)

Melody Aug 26, 2017

#3**+1 **

b-Melody: The total should be the coefficient of x^20 when the following is expanded:

expand | (x + x^2 + x^3 + x^4 + x^5 + x^6)^10

=x^60 + 10 x^59 + 55 x^58 + 220 x^57 + 715 x^56 + 2002 x^55 + 4995 x^54 + 11340 x^53 + 23760 x^52 + 46420 x^51 + 85228 x^50 + 147940 x^49 + 243925 x^48 + 383470 x^47 + 576565 x^46 + .................. + 383470 x^23 + 243925 x^22 + 147940 x^21 + **85228 x^20 **+ 46420 x^19 + 23760 x^18 + 11340 x^17 + 4995 x^16 + 2002 x^15 + 715 x^14 + 220 x^13 + 55 x^12 + 10 x^11 + x^10.

And the probability of this occurring is: 85,228 / 6^10 = 0.00140951.......

Similarly for a: expand | (x + x^2 + x^3 + x^4 + x^5 + x^6)^3

=x^18 + 3 x^17 + 6 x^16 + 10 x^15 + 15 x^14 + 21 x^13 + 25 x^12 + 27 x^11 + 27 x^10 +** 25 x^9** + 21 x^8 + 15 x^7 + 10 x^6 + 6 x^5 + 3 x^4 + x^3.

And the probability is: 25 / 6^3 =0.1157407..........

Guest Aug 26, 2017

edited by
Guest
Aug 26, 2017

#5**+1 **

I'm going through your calculations with a magnifying glass!, to see IF I can spot it!!. If I can find it, I will let you know.

Guest Aug 26, 2017

#8**+3 **

Are you a Nancy Drew fan, Mr. BB?

After you are done with your search, you can write some fan fiction: *The Case of the Missing Combinations.* By BB Eyes

I have one in progress: __The Case of the Missing Frontal Lobe __

My cat has one in progress too: *The Case of the Missing Monopoly Money *

He also just signed a contract with a publisher for a “How To” book. The title is *How to* *Embezzle* *Monopoly Money (and Get Away with it)*

I’ll send you a copy—one signed by my cat. Just post your Federal pen ID number on here.

GingerAle
Aug 26, 2017

#10**+1 **

Can I have your cats autograph to please Ginger. :)

Look what you have done to your poor cat!

Melody
Aug 26, 2017

#12**+3 **

I didn’t do that to him. D.C. did that to himself a few years ago when he was working on the * Mystery of The Stoned Hamsters*. A pet shop noticed the some of the hamsters were acting strange and hired D.C. to investigate. It didn’t take him long to find out the hamsters were visiting a medical marijuana dispensary two doors away from the pet store.

That’s a picture of D.C., taken a few hours after solving the case. To D.C., hamsters –stoned or not, are a delicacy; he ate a one, which gave him the munchies so he ate several more of them. He was stoned for days. It wasn’t too bad, except he kept reciting beatnik poetry.

GingerAle
Aug 28, 2017

#6**+1 **

There is a small mistake in columns 9, 10, 11. They should all be 840. Still looking!.

Guest Aug 26, 2017

#7**+1 **

OK. I think I found the mistake !!.

You missed 2 combinations: 5, 4, 1 =10!/5!4! =1260 and 5, 3, 2 =10!/5!3!2!=2520

So, 1260+2520+40(from column 10, 11) =3820 which is the difference between 85228 - 81408=3820.

Guest Aug 26, 2017

#11**0 **

I was thinking, this is like putting 9 indistinguishable balls into 3 distinguishable boxes where every box must have one ball. Once you put a ball in each box then we have two but 6 balls into 3 boxes. We can use stars (are the balls) and bars (the box dividers). For example this |**||****, would be 2-0-4. So that's the same as 9C3=84.

Jz1234 Aug 26, 2017

#15**+1 **

Hi Jz1234

It is not like that because the sum of the numbers must be 9.

So you cannot use a stars and bars method :)

Melody
Aug 28, 2017

#16**0 **

Hmm. His calculation was wrong, but the stars and bars method works perfectly when there is no upper bound (Im the same guest that wrote the last post about solving the question using inclusion and exlusion). The answer to the question he presented in THIS post (not in the original question) is 8C3 not 9C3 (because we need to use k-1 bars in order to divide this to k boxes, you used 3 bars to divide this to 3 boxes)

Guest Aug 28, 2017

#13**0 **

Sadly, melody's way of answering the questions is clumsy and complicated, and takes a lot of time when the numbers get larger (assuming you remember to count every combinations exactly once). This question is a famous one in combinatorics and has a much, MUCH simpler answer than the answers you gave. after reading Jz1234's last post i see he ALMOST got the right answer (and that is VERY impressive, well done!). jz1234's answers works only if we dont mention any upper bound in the question. Let me explain:

suppose we have r dices, and suppose every dice has a different color (the dices are distinguishable). The first dice is D_{1}, the second one is D_{2} and so on. suppose the i^{th} dice has S_{i} sides, and suppose on every side a different integer between 1 to S_{i} is written. We need to find the number of ways we can roll the dices to get the sum A.

Jz1234 Realized the number of different rolls is also the number of different solutions to the equation x_{1}+....+x_{r}=A where

0 i i+1. He also realized that the number of solutions to the equations is the number of solutions to the exuation x _{1}+....+x _{r}=A-r where 0= i= i-1. Now, Jz1234's solution holds when there is no upper bound S _{i} to any of the dices, and the solution he mentioned can be calculated easily. Sadly this is not the case for the general solution (with upper bounds).

suppose B_{i}- the set containing every combnations of the integers x_{1}, ...., x_{r} where x_{1}+....+x_{r}=A-r and x_{i}>S_{i}-1. Using the inclusion-exclusion principle we can get the solution (https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle).

As far as i know there is no general, nice formula that finds the number of rolls, but that is the best way i can think of for solving this question.

Guest Aug 28, 2017

#17**+3 **

**Guest all you gave was an example of arrogance squared.**

*Sadly, melody's way of answering the questions is clumsy and complicated, and takes a lot of time when the numbers get larger (assuming you remember to count every combinations exactly once). *

**This isn’t sad, nor is this clumsy.** It may be unwieldy because of its size, but it’s not clumsy. Complicated? Yes, of course, it is, and no matter how it’s answered it’s a complicated question. Melody’s presentation explains what is going on inside the formulas and algorithms used to solve this. Very un-advanced students are the target audience, here. They don’t know what generating function is. I doubt there are five members that know what a generating function is –let alone how do one.

*This question is a famous one in combinatorics and has a much, MUCH simpler answer than the answers you gave. after reading Jz1234's last post i see he ALMOST got the right answer (and that is VERY impressive, well done!). *

**No, he didn’t almost get the right answer. **

*jz1234's answers works only if we dont mention any upper bound in the question. *

No it doesn’t. If the goal is Mars, that formula doesn’t get you the moon.

*Let me explain:*

**[Sadly] Your explanation is laborious [Clumsy] and doesn’t produce a very usable result. **

*As far as i know there is no general, nice formula that finds the number of rolls, but that is the best way i can think of for solving this question*

**Well you don’t know much!** I know of a general, [nice] formula that gives the exact result. (Lancelot Link gave it to me when I was studying AOPS.)

\(\sum \limits_{k=0}^{m}\binom{n}{k} (-1)^k \binom{p – s*k – 1}{p – s*k – n} \leftarrow \text {where p is the point (sum) target, n is the # of die, }\\ \hspace {47mm} \text{s is the number of sides, and } m=\lfloor \dfrac{p-n}{s}\rfloor \\\)

This formula is from J. V. Uspensky’s Introduction to Mathematical Probability (1937). There is an extensive derivation here.

Here’s the Wolfram code for this fromula:

This is compliments of Lancelot Link. (I think that Chimp could program Wolfram to bring back a burning sample of brimstone from heII.)

The variables are preset. Paste this into Wolfram Alpha.

p=20,n=10,s=6; sum (binom(n,k)*(-1)^k*(binom((p-s*k-1),( p-s*k-n))) from k=0 to Floor[(p - n)/s])

Or click here.

**Now, take your [much, much] arrogant and useless commentary and shove it up your [sad, sad] áss! **

GingerAle
Aug 28, 2017

#18**0 **

I dont even know what to say ginger. I was not trying to be arrogant, nor i was trying to attack her or any other moderator. I thought the way she presented her answer was a bit complicated and hard to find. That's it. I already mentioned that there are easier ways to answer this, and i thought that maybe presenting JUST the table with all the combinations is complicated for that. Im sure melody can take a little bit of criticism.

You also corrected me about "this is the best way i know"-

1. I was talking about the general case where every die has a different number of sides.

2. This is the best way *i* know (knew).

3. When i said "simple" i meant a single binomial coefficient that wi solve the question like the simple binomial coefficient that is the solution when there's no upper bound.

(But be my guest, find a general, simple formula for the general case, it's worth trying isnt it ;))

________________

It seems as if you're mostly trying to defend melody, whenever someone says something about her that you dont like. I didnt say melody is a bad moderator, nor that she's stupid for not using the way i did. I pointed out the fact that there is a simpler solution that i THOUGHT she should've presented.

But when im complimenting jz1234 for coming up with an idea that was closer than everyone else in the thread? (Again, this might seem like im attacking melody. You can calm down, ginger! Take a deep breath, drink a glass of water, because im not attacking her, im mentioning a fact)- no, dont even THINK about complimenting him, because he didng give the EXACT formula for the solution.

Now, you may not believe me, but i knew you'll criticize me for my post. I knew you'll call me a blarney banker, i knew you'll defend melody. This leaves two options:

1.Im the oracle!

2. You're just getting predictable.

tl;dr: grow up

ThePuppetMaster Aug 29, 2017

edited by
Guest
Aug 29, 2017

#20**+3 **

^{I dont even know what to say ginger. }

Well, you sure do say a lot for someone who doesn’t know what to say. Are you just shooting off words in the dark hoping some of them find the light of day?

^{I was not trying to be arrogant, nor i was trying to attack her or any other moderator. }

Maybe not, but just because you are unaware of your arrogance doesn’t mean you are not arrogant.

^{I thought the way she presented her answer was a bit complicated and hard to find. That's it. }

Critiquing a presentation is fine. But your choice of words (“sad,” “much, much”) creates an artificial hyper-emotional atmosphere that’s conducive for dismissal of a presentation that has a high potential for explaining the iterative nuances of a complex oscillating combinatorics problem to any student that would want to learn from it. Only a dedicated teacher would spend the time to do this. These are labor intensive and time consuming to do manually on a spreadsheet –I know, I’ve done them.

Heureka made a similar presentation for the exact same question, here, by writing a C++ program. I expect his program parsed and sorted these in a fraction of a second. I don’t know how long it took him to write and debug the program—an hour maybe, but I’m sure it was only a fraction of the time Melody spent on her presentation. Heureka’s presentation was well received—as it should have been, by Sir CPhill, Mr. BB –the Blarney Banker (amazing) and others who didn’t comment. Melody’s presentation is well received too. Mr. BB was intrigued enough to search for and find the missing combinations. I do hope he writes about it in *The Case of the Missing Combinations. *

^{Im sure melody can take a little bit of criticism.}

I’m sure she can take a banana-boatload of it. She’s a teacher –a successful one, and that means her skin has to be as thick as rhinoceros hide. How many parents have blamed her because their children are too stupid to learn?

“My Johnny did good at math, until he was in your class!”

“Buffy doesn’t understand her homework and I can’t explain it to her, which means you did a crappy job teaching her. Let me tell you how to do your job.”

The question isn’t whether Melody can take a little bit of criticism, it’s whether her post is worthy of it. It wasn’t. It most defiantly wasn’t when contrasted with your alternative method for a solution. A method so complex, that there probably isn’t one person on here that knows how to use it—that includes you.

^{But when im complimenting jz1234 for coming up with an idea that was closer than everyone else in the thread? (Again, this might seem like im attacking melody. You can calm down, ginger! Take a deep breath, drink a glass of water, because im not attacking her, im mentioning a fact)- no, dont even THINK about complimenting him, because he didng give the EXACT formula for the solution.}

I didn’t think you were attacking Melody, I just think you are full of blarney! (Blarney is a euphemism for Bullshít.) You blòw blarney smoke like it has substance. You seem to think that if you say something with a certainty that makes it true. Congratulations to JZ for thinking along binomial lines for a solution—he is defiantly thinking in the right direction.

However, you are not thinking in the right direction. Saying that formula is very close to correct and that it works for unbounded ceilings is blarney.

^{He also realized that the number of solutions to the equations is the number of solutions to the exuation x 1+....+x r=A-r where 0= i= i-1. Now, Jz1234's solution holds when there is no upper bound S i to any of the dices,}

Your “proof” is blarney: you present a plop of incomprehensible ASCII slop but fail to present a final formula. The Stars and Bars method is among the most simple of combinatoric formulas to use and present. Why didn’t you demonstrate it after your complex presentation? The reason is simple: you are full of ~~blarney~~ Bullshít!

**I said your** **commentary was arrogant and useless, and it is, but your math presentation is so delusional and chaotic, it indicates a serious mental illness. The best part of your presentation is “ 0= i= i-1” and that, by itself, is truly worthy of a blòódy Field’s medal!**

You do have great mindreading skills though, else how would you know what Jz1234 realized? Of course, maybe he’s one of your puppets, PM. How many do you have on here?

^{Now, you may not believe me, but i knew you'll criticize me for my post. }

I believe you. You have demonstrated advanced mind reading skills, so predicting the future isn’t much more advanced.

^{I knew you'll call me a blarney banker,}

Ah, you prognosticated this one wrong. I only address the Blarney Banker as such, and he’s not one of your puppets, so you shouldn’t take it personally.

^{ i knew you'll defend melody. }

Yep! I don’t really need to though; she’s quite capable of defending herself. There’s an instinct in us Chimps to protect and warn the troop when any kind of threat comes near. Your toxic offal fits the bill.

^{This leaves two options:}

^{1.Im the oracle!}

Did you mean orifice? If you did, then Ok.

I think you are most likely TPM. Did you know (via your mind reading ability) that one of my favorite piano pieces is __Funeral March of a Marionette__? Rather fitting, I think.

https://youtu.be/gkjNeHlSoko?t=45

^{2. You're just getting predictable.}

You are partly right. I’ve been predictable for a long time. Most of us genetically enhanced chimps are predictable, except when we are not.

^{ tl;dr: }

I’m not surprised; you do seem lazy.

^{grow up}

Never!

Until next time, don’t let wagging heads, jointless gestures, and long monologues wear you out.

GA

GingerAle Aug 30, 2017

#21**+1 **

LMFAO! Ginger Ale, you truly are remarkable! You post some interesting and impressive math, and you know how to turn a phrase in humor to bite or compliment, or just to entertain. Frequently, it’s all three at the same time. You make a great banana daiquiri too.

How did you know it was *The Puppet Master*? There are not many posts by him for you to use as a comparison. I didn’t know who you meant by TPM until I saw that he signed on and claimed his copyright. I’d bet this was unintentional. The puppets pulled his strings this time. hahahaha!

Your __Funeral March For a Marionette__ reference is brilliant and so is your valediction, “...don’t let wagging heads, jointless gestures, and long monologues wear you out.”

I’ll be laughing for days!

JacobBernoulli
Aug 30, 2017