A challenge my friend gave me

[pineapplewithmouse]

There is a cube with 6 equal sides. You roll the cube 4 times. How much permutations of possible numbers can be, if it's known that the sum of the 4 numbers in the 4 rolls is equal to 18?
I tried to make a 6*6 chart with the numbers 1-6 on each side, but not all the possible combinations were there.
Someone else (I am not sure) solved it with some type of distribution, but the friend who asked the question said that this is not an efficient way.
Maybe it's out of my knowledge in math, or maybe I need some creative thinking, but I lack of this .
Can someone help me?

 

[Dr.Peterson]

There is a cube with 6 equal sides. You roll the cube 4 times. How much permutations of possible numbers can be, if it's known that the sum of the 4 numbers in the 4 rolls is equal to 18?
I tried to make a 6*6 chart with the numbers 1-6 on each side, but not all the possible combinations were there.
Someone else (I am not sure) solved it with some type of distribution, but the friend who asked the question said that this is not an efficient way.
Maybe it's out of my knowledge in math, or maybe I need some creative thinking, but I lack of this .
Can someone help me?

The wording is extremely unclear! It doesn't even mention what the numbers are, I assume the cube is a die with the numbers 1 through 6 on the faces.

I think it's just asking for the number of different sequences of four numbers from 1 to 6, with repetition allowed, have a sum of 18.

A 6x6 chart would only cover pairs of numbers. If you want to make a list, you might start with the largest numbers first: 6,6,6,0 isn't allowed, since there is no 0; so the first in your list might be 6,6,5,1, then 6,6,4,2 and 6,6,3,3, and so on. Try doing that, and as you go, be looking for shortcuts. Listing will certainly not be efficient, but it can lead to insights if you pay attention.

If you want a more "mathematical" method, we'll need to know what methods you have learned. The fastest methods (if there are any) will probably be beyond you, based on your description of yourself.

 

[Steven G]

Continuing with what Dr Peterson said...

6651
6642
6633
6624
6615
6561
6552
...
6462
6453
...
4266
..
3663
3654
...
3366
...
1665
...

 

[Cubist]

The wording is extremely unclear! It doesn't even mention what the numbers are,

I agree. This wording makes me very suspicious

@pineapplewithmouse if you can choose your own numbers to write on the cube then the problem becomes much simpler. Can you see why? It might be worth checking this with your friend.

If it is a standard die, then there's a very fast way using inclusion-exclusion. But first it would be good to see what you can do with the suggested list - and can you see how to avoid listing the whole lot?

 

[pka]

If as others have said that we are forced to assume that this number cube is an ordinary die.
Tossing four dice or one die four time yields same outcomes. [imath] {\left( {\sum\limits_{k = 1}^6 {{x^k}} } \right)^4}=x^{24} + 4 x^{23} + 10 x^{22} + 20 x^{21}+ 35 x^{20}+ 56 x^{19}+ 80 x^{18}+ 104 x^{17}+ 125 x^{16} + 140 x^{15} +146 x^{14}+ 140 x^{13} + 125 x^{12}+ 104 x^{11} + 80 x^{10}+ 56 x^9 + 35 x^8 + 20 x^7 + 10 x^6 + 4 x^5 + x^4[/imath]
That is known as a generating polynominal the term [imath]104x^{17}[/imath] tells us that there one hundred and four ways to get seventeen tossing four dice.

 

[pineapplewithmouse]

Oh I am very sorry, yes, the cube is a die with the numbers 1-6.
About my math level, let's say I didn't learn calculus yet.

 

[Deleted member 4993]

Oh I am very sorry, yes, the cube is a die with the numbers 1-6.
About my math level, let's say I didn't learn calculus yet.

You DO NOT NEED calculus to solve this problem.

Read response #6 very carefully. The answer is already given there.

 

[pka]

Oh I am very sorry, yes, the cube is a die with the numbers 1-6.
About my math level, let's say I didn't learn calculus yet.

No need to worry. And as far as not having studied Calculus that is not a problem.
The expanded expression I gave you was so you can find the number of ways to any sum by tossing [imath]n[/imath] dice.
The exponent on [imath]x[/imath] the sum while the coefficient is the number of ways.
If you can use WolframAlpha to expand [imath]{\left( {\sum\limits_{k = 1}^6 {{x^k}} } \right)^n}[/imath]
you can find the no of eays ways for any [imath]\bf n[/imath] dice.

[imath][/imath][imath][/imath]

 

[Cubist]

If it is a standard die, then there's a very fast way using inclusion-exclusion.

I'll describe an alternative method (now that an answer is mostly given above). It's based on "stars and bars". The 4 buckets each represent the number obtained on the throw of a die. The minimum throw is "1", therefore we must get 14 extra pips distributed in some way between the 4 throws in order to obtain a total of 18. But we need to exclude any ways that would give a single bucket more than 5 pips, that's why the inclusion-exclusion is needed...

the number of ways to rearrange |||************** is 680
minus the ways to rearrange |||******** times [imath]\binom{4}{1}[/imath] = 660
plus ways to rearrange |||** times [imath]\binom{4}{2}[/imath] = 60
680 - 660 + 60 = ?? ways

This is more difficult to understand than the expansion of a generating poly. However, it's probably quicker to compute.

@pineapplewithmouse have you tried to make the suggested list?

Spoiler: list hint

You only need 8 lines in the list if you have an extra column for the number of ways that each line can be permuted
Write each result so that subsequent digits are descending

result : permutations
6651 : 4! / (2!) = 12
6642 : ?
6633 : 4! / (2! * 2!) = 6
6552 : ?
... 4 more lines
The 8 permutation numbers will sum to the correct answer

 

[pineapplewithmouse]

I am sorry @Cubist I don't really understand you way
@pka I checked in the internet about the sigma operation, and I have a question:
Why it works?
I mean, I understand what each number/variable means (well, maybe not the x)
But why it makes that the coefficient of x is the number of ways you can get the exponent of x after the number you put outside the brackets?

 

[Cubist]

I am sorry @Cubist I don't really understand you way

Please ignore the first method in my post.

However, you should be able to understand how to list out the possible throws. Here's the spoiler from my post above...

You only need 8 lines in the list if you have an extra column for the number of ways that each line can be permuted
Write each result so that subsequent digits are descending

result : permutations
6651 : 4! / (2!) = 12
6642 : ?
6633 : 4! / (2! * 2!) = 6
6552 : ?
... 4 more lines
The 8 permutation numbers will sum to the correct answer

All of the digits in the left hand column represent the 4 throws of a die. These digits sum to 18. For example 6633 implies throwing 6 then 6 then 3 then 3, and 6+6+3+3 = 18

In the list above, the 6633 can be rearranged in the following 6 ways {3366, 3636, 3663, 6336, 6363, 6633}. And 6 is the number I'm calculating in the "permutations" column. Another row has 6651 and this can be permuted 12 ways {1566, 1656, 1665, 5166, 5616, 5661, 6156, 6165, 6516, 6561, 6615, 6651}.

Regarding calculation of the permutations, the word "MISSISSIPPI" should make it obvious how the calculation works. There are 11 letters in total, 4 letter S, 4 letter I, and 2 letter P

[math]\frac{11!}{4! \times 4! \times 2!}=34,650[/math]{IIIIMPPSSSS, IIIIMPSPSSS, IIIIMPSSPSS, ... MISSISSIPPI ... , SSSSPPIIMII, SSSSPPIMIII, SSSSPPMIIII}

This makes it quick to list all the possible outcomes, because you can ignore all the permutations.

Do you understand this, and if so can you complete the list that I started?

 

[pineapplewithmouse]

I got 68

6651 : 12
6642 : 12
6633 : 6
6552 : 12
5553 : 4
5544 : 6
5562 : 12
4464 : 4

12+12+6+12+6+12+4+4=68

But if that's true, why @pka way didn't work?

 

[pka]

I got 68
What does 68 count?
@pka way didn't work?

If you are counting the number of ways that the sum of [imath]18[/imath] appears when four standard dice are tossed, then it is incorrect.
The correct answer is [imath]\bf 80[/imath].Your attempt at listing all of the ways is difficult at best
In reply #5, I gave you an way to find the number of ways to roll a sum of [imath]n=4\text{ to }24.[/imath].
To find that number look at the expansiion for the term [imath]J x^n[/imath]
the [imath]J[/imath](coefficent) tells you that there [imath]J[/imath] ways to roll the sum [imath]\bf n~[/imath](the exponent)
As for your being able to reproduce such an expansion, no one suggested that.


[imath][/imath][imath][/imath][imath][/imath]

 

[Cubist]

I got 68

6651 : 12
6642 : 12
6633 : 6
6552 : 12
5553 : 4
5544 : 6
5562 : 12
4464 : 4

Well done . You just made two mistakes...

6651 : 12 ✔
6642 : 12 ✔
6633 : 6 ✔
6552 : 12 ✔
6543 : 24 this entry was missing in your attempt
6444 : 4 ✔ (this is your 4464 line but with the digits in descending order)
5553 : 4 ✔
5544 : 6 ✔
You also had a line "5562 : 12" (but this is actually part of the 6552 line, since the digits here must be in descending order)

12+12+6+12+24+4+4+6 = 80

 

[pineapplewithmouse]

Well done . You just made two mistakes...

6651 : 12 ✔
6642 : 12 ✔
6633 : 6 ✔
6552 : 12 ✔
6543 : 24 this entry was missing in your attempt
6444 : 4 ✔ (this is your 4464 line but with the digits in descending order)
5553 : 4 ✔
5544 : 6 ✔
You also had a line "5562 : 12" (but this is actually part of the 6552 line, since the digits here must be in descending order)

12+12+6+12+24+4+4+6 = 80

Oh I wrote the 6552 twice. So it is 80.

If you are counting the number of ways that the sum of [imath]18[/imath] appears when four standard dice are tossed, then it is incorrect.
The correct answer is [imath]\bf 80[/imath].Your attempt at listing all of the ways is difficult at best
In reply #5, I gave you an way to find the number of ways to roll a sum of [imath]n=4\text{ to }24.[/imath].
To find that number look at the expansiion for the term [imath]J x^n[/imath]
the [imath]J[/imath](coefficent) tells you that there [imath]J[/imath] ways to roll the sum [imath]\bf n~[/imath](the exponent)
As for your being able to reproduce such an expansion, no one suggested that.


[imath][/imath][imath][/imath][imath][/imath]

I am sorry, I just said to my friend that the answer is 80 and he said it's wrong so I thought your way is wrong.
But even with the list method it's 80, so I guess it's the real answer.

Thanks @pka and @Cubist for helping me

 

[Steven G]

I mean, I understand what each number/variable means (well, maybe not the x)
But why it makes that the coefficient of x is the number of ways you can get the exponent of x after the number you put outside the brackets?

Please multiply out (x¹+x²+x³+x⁴+x⁵+x⁶)².
Now make the 6x6 table you were talking about earlier. Do you see, for example, that the coefficient of x⁵ is 4. That is the number of ways to roll two dice and get a 5.
Now why does this work? Think about how you get x⁵ terms when you multiply out the two factors.