Given that $1 \leq i,j \leq 6$ and that $1 \leq k \leq 12$, how many solutions there are given that $i+j>k$.

The problem appears when we have 2 players A and B. A using 2 dice (1 to 6 ) and B using one die (1 to 12). If the sum of the dice from player A respects $i+j>k$ then A won. If $i+j=k$ no one owns.

I've resolve it hard style, enumerating all the cases (from the 36 cases possibles for the sum of A) and I've found the answer. My problem is how can I use a more general rule for it.

I know I will have to use the stars-and-bars but I can't figure out out.

Any help/hint will be appreciated!


closed as off-topic by Mike Pierce, Shailesh, Daniel W. Farlow, JonMark Perry, user296602 Apr 14 '16 at 3:53

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  • $\begingroup$ There are $6 \cdot 6 = 36$ ordered pairs $(i, j)$ with $1 \leq i, j \leq 6$. $2 \leq i + j \leq 12$. Count how many ordered pairs yield each sum. $\endgroup$ – N. F. Taussig Apr 14 '16 at 0:57

The star-and-bars theorem says for any pair of positive integers n and k, the number of k-tuples of positive integers whose sum is n is equal to the number of (k − 1)-element subsets of a set with n − 1 elements. However in your case, I don't think using star-and-bars theorem is a good idea, because you have a restriction on i and j. So in this case, I would suggest counting the solutions when k equals different values and try to find a pattern.

I'm using (i,j) to denote value of i and j.

k=1: all possible combinations of i and j works, so that would be 6*6=36.

k=2: all possible combinations except (1,1), so that would be 36-1=35.

k=3: all possible combinations except (1,1),(1,2),(2,1), so that would be 36-3=33.


k=10: (5,6),(6,5),(6,6), 3 total ways

k=11: (6,6), 1 way

k=12: no combination of (i,j) satisfies.

We noticed that if we group k=1 and k=12, k=2 and k=11, k=3 and k=10 etc., each group has a 36 possible ways to choose i and j. We have 6 groups in total. Therefore, we have 36*6=216 possible solutions to the equation.

  • $\begingroup$ Hey! Thanks. I've done that. My problem (and sorry for not explaining it better on the question) is if there is any way to extract a more general thing from it. For example this case is simple because we have 1<= k <= 12 and i+j<k. But what if we had i,j,n ? $\endgroup$ – Bakayarou Apr 14 '16 at 15:54

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