Thursday, September 27, 2007

How Many Ways Can You Get to 100?

How many strictly positive integer solutions (x, y, z) are there, such that x + y + z = 100


  1. my unofficial answer will be 5,151 combinations. Here's my thinking:

    where x=0, y+z=100 | 101 combinations

    where x=1, y+z=99 | 100 combinations

    where x=2, y+z=98 | 99 combinations

    therefore, x=n, y+z=101-n | 101-n combinations.

    sum(x) = (101+1)(101/2)= 5151

    I think that (but dont have the time to check) covers all the combinations because for each value of x, every possible value of y+z is calculated, therefore for every value of y and z, x+y and x+z = 100 respectively.

    please correct me if my logic is wrong.

  2. The answer is 4851 ways. It helps if you think of putting 100 stones into three piles.

    For a technical answer, read on:

    By removing one stone from each pile, this is the number of ways you can arrange m-k identical stones into k (possibly empty) piles.

    Now, view the k piles as a numbered set .

    Write on each stone the number of a chosen pile and order the stones accordingly.

    The numbered stones constitute a combination with repetition of k elements (the numbers) choose m-k (the stones). The number of combinations can be calculated as:
    In our case, m = 100 and k = 3, so we have 99!/97!*2! = 4851

    Although, I could be wrong. ;-)

  3. well technically o is not positive or negative so u can not figure one of the numbers as zero and do decimals count? because if so then the answer would b infinite

  4. Hi anonymous, I'm not sure I understand your question, so forgive me. But if each of the x, y and z need to be strictly positive integers, then 0 is out and so are decimals.

  5. "By removing one stone from each pile, this is the number of ways you can arrange m-k identical stones into k (possibly empty) piles."

    Hey there, this is a different annonymous user but when you say "possibly empty" is that erroneous considering you are leaving 0 out of the question. (Strictly positive integers)


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