Number of Combinations for 7 DiceDate: 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 |
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