Number of Combinations for 7 Dice

Date: 22 Feb 1995 01:22:08 -0500
From: Michel Bertler
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 order 
          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.Bertler@Deltacom.cam.org

Date: 27 Feb 1995 15:57:05 -0500
From: Stephen Weimar
Subject: Re: 7 dice # of combinations?

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.

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