CombinatorialCase110: Difference between revisions
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3 2 1 | 3 2 1 | ||
⍝ Permutations of length L of R items | ⍝ Permutations of length L of R items | ||
⍝ Labeled balls & boxes, ≤1 # | ⍝ Labeled balls & boxes, ≤1 # Balls per Box | ||
!3 | !3 | ||
6 | 6 |
Revision as of 20:18, 29 April 2017
This case produces L-Permutations (also called Partial Permutations or Sequences Without Repetition) of R items, where when L=R produces the familiar permutations !R. The length of each permutation returned is always L.
- L labeled balls (1), R labeled boxes (1), at most one ball per box (0)
- Sensitive to ⎕IO
- Counted result is an integer scalar
- Generated result is an integer matrix.
The count for this function is (!⍠(-L))R where (!⍠L)R calculates the Rising and Falling Factorial.
For example:
If we have 3 labeled balls (❶❷❸) and 3 labeled boxes (123) with at most one ball per box, there are 6 (↔ (!⍠¯3)3 ↔ 3×2×1) ways to meet these criteria:
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The diagram above corresponds to
110 1‼3 3 1 2 3 2 1 3 2 3 1 1 3 2 3 1 2 3 2 1 ⍝ Permutations of length L of R items ⍝ Labeled balls & boxes, ≤1 # Balls per Box !3 6 110‼3 3 6 110 0‼3 3 6 ⍴110 1‼3 3 6 3 110 1‼3 3 1 2 3 2 1 3 2 3 1 1 3 2 3 1 2 3 2 1 110 1‼2 3 1 2 2 1 1 3 3 1 2 3 3 2 110 1‼1 3 1 2 3
A function to calculate the permutations of R items could be defined as
perm←{110 1‼⍵ ⍵}