Number of Combinations for 7 Dice

Date: 22 Feb 1995 01:22:08 -0500
From: Michel
Subject: 7 dice # of combinations?

Hello there!

I'm looking for the derived formula solution for the following problem
someone posted in a math echo and to whom I've replied this:

  There are seven dice.  Each die has six faces.  How many different
  combinations are there of these seven dice?  In other words, 1112222
  counts as the same combination as 2222111 or 1222211, etc.

The total maximum outcomes from 7 dice are:  6^7 = 279936

As you can see, it would be quite exhaustive to build an array to sort
them all out!  But let's use a more simplistic 2 dice pattern (a black 
and a red dice):

                       Red die

             | 1   2   3   4   5   6
            --+------------------------
            1| 2   3   4   5   6   7
         B   |
         l  2| 3   4   5   6   7   8
         a   |
         c  3| 4   5   6   7   8   9
         k   |
            4| 5   6   7   8   9  10
         d   |
         i  5| 6   7   8   9  10  11
         e   |
            6| 7   8   9  10  11  12

The above table is to show all 36 possible outcomes from 2 dice (6^2).
Now those 36 are to be considered as permutations let's extract the
combinations from them:

1,1
1,2 2,2
1,3 2,3 3,3
1,4 2,4 3,4 4,4
1,5 2,5 3,5 4,5 5,5
1,6 2,6 3,6 4,6 5,6 6,6

Here as we can see, there are 21 combinations to be extracted from 2 
dice.

6+5+4+3+2+1 or 6(6+1)/2 = 21

But let's go further and try with a 3 dice pattern:

1,1,1
1,1,2
1,1,3
1,1,4
1,1,5
1,1,6
1,2,2 2,2,2
1,2,3 2,2,3
1,2,4 2,2,4
1,2,5 2,2,5
1,2,6 2,2,6
1,3,3 2,3,3 3,3,3
1,3,4 2,3,4 3,3,4
1,3,5 2,3,5 3,3,5
1,3,6 2,3,6 3,3,6
1,4,4 2,4,4 3,4,4 4,4,4
1,4,5 2,4,5 3,4,5 4,4,5
1,4,6 2,4,6 3,4,6 4,4,6
1,5,5 2,5,5 3,5,5 4,5,5 5,5,5
1,5,6 2,5,6 3,5,6 4,5,6 5,5,6
1,6,6 2,6,6 3,6,6 4,6,6 5,6,6 6,6,6

Here we get 56 combinations:

21+15+10+6+3+1 = 56

Obviously, n(n+1)/2 doesn't work anymore because of the supplementary 
dice correlation, things are probably getting worse with 4, 5, 6 and 7 
dice and you'll be on your own to experiment.  I wish I could give you any 
more clues, but the few books I have here on probability & statistics 
have not been so far very specific on this.

There must be a way to find out exactly how many possible combinations
we could get from any number of dice, so far I have not found it!  Could
anyone provide the answer?

Thanks!

Michel

Date: 27 Feb 1995 15:57:05 -0500
From: Dr. Steve
Subject: Re: 7 dice # of combinations?

Hi Michel,

Here is a way to get the formula you asked for.  Let's say there
are N dice in all (so your question sets N equal to 7).  As you
have done, I will list the combinations in increasing order, but
to make things a little clearer, I will put an X every time you
are finished listing all of one kind of number and go on to the
next.  For example, instead of

                1 1 2 3 5 5 5

I will write

                1 1 X 2 X 3 X X 5 5 5 X  .

Notice there are 2 X's after the 3, or rather that there are
zero 4's between the third and fourth X.  So now the list of
combinations begins like this:

                1 1 1 1 1 1 1 X X X X X

                1 1 1 1 1 1 X 2 X X X X

                1 1 1 1 1 1 X X 3 X X X

                1 1 1 1 1 1 X X X 4 X X

                1 1 1 1 1 1 X X X X 5 X

                1 1 1 1 1 1 X X X X X 6

                1 1 1 1 1 X 2 2 X X X X

                1 1 1 1 1 X 2 X 3 X X X

and so on.  The useful thing about writing it this way is that
you can tell what the combination is merely by knowing the 
positions of the five X's.  In other words, there is a 1 to 1 
correspondence between combinations of N dice and ways of 
choosing 5 places to put an X out of a total of N+5 positions.  
(Do you see why there are N+5 positions?  In the above examples, 
7 of the positions are filled by numbers and 5 get X's .)  Now the 
problem is easily solved:  it is the binomial coefficient "N+5 
choose 5", which is equal to:

        (N+5) (N+4) (N+3) (N+2) (N+1) / 120 .

So for N=1 you get 6, for N=2 you get 21, for N=3 you get 56,
and so on, in accordance with your observations.

I hope you find this explanation comprehensible, but if not,
don't hesitate to ask specific follow-up questions.

- Doctor Steve, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/15/2009 at 00:19:19
From: Dennis
Subject: how many unique combinations for N dice

I have studied Dr. Math's answer to "Number of Combinations for 7 
Dice".  I understand the mechanics of the process described and the 
binomial coefficient reasoning, but I cannot figure out why the 
number of entries in each line of the sample table is N+5.  Dr. Math 
says rhetorically "Do you see why there are N+5 positions?" My 
answer is no I do not.  Can you help?

Why is it N+5?  I can deduce N+5 by creating a table and seeing that 
it works for N = 1, 2 and 3. In other words I can see this works by 
empirically doing the table and counting.  But I cannot deduce the 
general rule.

I can solve a specific N-dice problem pretty easily using EXCEL.  I 
just let the dice faces be prime numbers instead of 1-6 and use the 
product of numbers of each possible roll to find combinations.  The 
products of all like combinations are all equal.  I can generate the 
table of all possible dice rolls using the EXCEL modulo function and 
sort and use a few numerical tricks to extract the desired result.  I 
have done this for N = 2 and 3.  I could do N = 4 pretty easily but 
after that it gets unwieldy.  Crude but effective.

Date: 05/15/2009 at 08:24:02
From: Doctor Anthony
Subject: Re: how many unique combinations for N dice

The problem, if I understand you correctly, is how many different, 
distinct, combinations of numbers you can get when you roll 7 dice. 
E.G. you could get the following set of 7 digits  1,1,2,3,4,5,5 with 
order not being important, so I have given them in ascending order.

You can model this as distributing 7 balls into 6 boxes with the 
boxes labeled 1 to 6.  See the diagram below:

    1  2   3   4   5   6
  |**|   |***|   |   | ** |    This shows  1,1,3,3,3,6,6

The diagram must start and end with a '|' but otherwise the 7 *'s and 
5 '|'s  can appear in any order.  That is a total of 7+5 = 12 objects 
with 7 alike of one kind and 5 alike of a second kind.

                                            12!
Total number of different arrangements  = -------  =  C(12,5)
                                           7! 5!

                                        =  792  <-----

The general formula with n dice would be  C(n+5,5)

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/15/2009 at 10:26:40
From: Dennis
Subject: Thank you (how many unique combinations for N dice)

Dr. Anthony,

Thank you for your very prompt and clear response.  I am a 70-year 
old tutor/volunteer at a local charter high school and mostly help in 
the advanced science and math courses with the math being traditional 
algebra/trig/calculus.  But the "why" of this problem eluded me.

This site is terrific.  Thanks again.

- Dennis