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Home/ Questions/Q 7655439
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T12:30:57+00:00 2026-05-31T12:30:57+00:00

So I’ve been working on a problem in my spare time and I’m stuck.

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So I’ve been working on a problem in my spare time and I’m stuck. Here is where I’m at. I have a number 40. It represents players. I’ve been given other numbers 39, 38, …. 10. These represent the scores of the first 30 players (1 -30). The rest of the players (31-40) have some unknown score. What I would like to do is find how many combinations of the scores are consistent with the given data.

So for a simpler example: if you have 3 players. One has a score of 1. Then the number of possible combinations of the scores is 3 (0,2; 2,0; 1,1), where (a,b) stands for the number of wins for player one and player two, respectively. A combination of (3,0) wouldn’t work because no person can have 3 wins. Nor would (0,0) work because we need a total of 3 wins (and wouldn’t get it with 0,0).

I’ve found the total possible number of games. This is the total number of games played, which means it is the total number of wins. (There are no ties.) Finally, I have a variable for the max wins per player (which is one less than the total number of players. No player can have more than that.)

I’ve tried finding the number of unique combinations by spreading out N wins to each player and then subtracting combinations that don’t fit the criteria. E.g., to figure out many ways to give 10 victories to 5 people with no more than 4 victories to each person, you would use:
C(14,4) – C(5,1)*C(9,4) + C(5,2)*C(4,4) = 381. C(14,4) comes from the formula C(n+k-1, k-1) (google bars and strips, I believe). The next is picking off the the ones with the 5 (not allowed), but adding in the ones we subtracted twice.

Yeah, there has got to be an easier way. Lastly, the numbers get so big that I’m not sure that my computer can adequately handle them. We’re talking about C(780, 39), which is 1.15495183 × 10^66. Regardless, there should be a better way of doing this.

To recap, you have 40 people. The scores of the first 30 people are 10 – 39. The last ten people have unknown scores. How many scores can you generate that are meet the criteria: all the scores add up to total possible wins and each player gets no more 39 wins.

Thoughts?

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  1. Editorial Team
    Editorial Team
    2026-05-31T12:30:58+00:00Added an answer on May 31, 2026 at 12:30 pm

    Generating functions:

    Since the question is more about math, but still on a programming QA site, let me give you a partial solution that works for many of these problems using a symbolic algebra (like Maple of Mathematica). I highly recommend that you grab an intro combinatorics book, these kind of questions are answered there.

    First of all the first 30 players who score 10-39 (with a total score of 735) are a bit of a red herring – what we would like to do is solve the other problem, the remaining 10 players whose score could be in the range of (0…39).

    If we think of the possible scores of the players as the polynomial:

    f(x) = x^0 + x^1 + x^2 + ... x^39
    

    Where a value of x^2 is the score of 2 for example, consider what this looks like

    f(x)^10    
    

    This represents the combined score of all 10 players, ie. the coefficent of x^385 is 2002, which represents the fact that there are 2002 ways for the 10 players to score 385. Wolfram Alpha (a programming lanuage IMO) can evaluate this for us.

    If you’d like to know how many possible ways of doing this, just substitute in x=1 in the expression giving 8,140,406,085,191,601, which just happens to be 39^10 (no surprise!)

    Why is this useful?

    While I know it may seem silly to some to set up all this machinery for a simple problem that can be solved on paper – the approach of generating functions is useful when the problem gets messy (and asymptotic analysis is possible). Consider the same problem, but now we restrict the players to only score prime numbers (2,3,5,7,11,…). How many possible ways can the 10 of them score a specific number, say 344? Just modify your f(x):

    f(x) = x^2 + x^3 + x^5 + x^7 + x^11 ...
    

    and repeat the process! (I get [x^344]f(x)^10 = 1390).

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