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I'm trying to figure out what the equation is that can calculate the total number of combinations for placing "pips" into rows, where each row has a maximum. I'm not quite sure if that explains it, so here's an example:

6 pips with 3 rows each with a maximum of 5, has 25 combinations:

  • "xxxxx" "x" ""
  • "xxxxx" "" "x"
  • "xxxx" "x" "x"
  • "xxxx" "xx" ""
  • "xxxx" "" "xx"
  • "xxx" "xxx" ""
  • "xxx" "" "xxx"
  • "xxx" "xx" "x"
  • "xxx" "x" "xx"
  • "xx" "xxxx" ""
  • "xx" "xxx" "x"
  • "xx" "xx" "xx"
  • "xx" "x" "xxx"
  • "xx" "" "xxxx"
  • "x" "xxxxx" ""
  • "x" "" "xxxxx"
  • "x" "xxxx" "x"
  • "x" "x" "xxxx"
  • "x" "xxx" "xx"
  • "x" "xx" "xxx"
  • "" "xxxxx" "x"
  • "" "x" "xxxxx"
  • "" "xxxx" "xx"
  • "" "xx" "xxxx"
  • "" "xxx" "xxx"

What's the equation that can calculate this, given a(m, r, p) = ? where m is the maximum number of pips per row, r is the number of rows, and p is the pips to select?

user270555
  • 3
June Rhodes
  • 103

2 Answers2

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Let us assume that the number of "pips" is p, there are r "baskets" to place those "pips" and the maximum number of "pips" that can be placed in each basket is m. Then a(m,r,p)=(p+r-1)!/(r-1)!p!- Σr(n-i) for i in {0,1,..,p-m} For the first part of the equation assume that you put all the p "pips" in line and you have to place r-1 lines, so that there are r baskets. Assumed that you have to choose the r-1 lines of p+r-1 places. so you must calculate the combination of p+r-1 things r-1 at a time. For the second part, because you have set a maximum number m of pips that can be placed in baskets you have to substract Σr(n-i) from the previous combination, because this is the number of not allowed combinations

Frentos
  • 3,041
Anastasia Foudouli
  • 1
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This answer has a recurrence that solves the problem. Your $a(m, r, p)$ is equivalent to $a_m(r-1, p)$ in that answer. Note that when coding an algorithm, you'll need a version of the binomial coefficient that yields $0$ for out-of-range values.

Frentos
  • 3,041