CombinatorialCase110: Difference between revisions
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This case produces '''<apll> | This case produces '''<apll>M</apll>-Permutations of <apll>N</apll> items''' (also called '''Partial Permutations''' or '''Sequences Without Repetition'''), where when <apll>M=N</apll> produces the familiar permutations <apll>!N</apll>. The length of each permutation returned is always <apll>M</apll>. | ||
* <apll> | * <apll>M</apll> labeled balls (1), <apll>N</apll> labeled boxes (1), at most one ball per box (0) | ||
* Sensitive to <apll>⎕IO</apll> | * Sensitive to <apll>⎕IO</apll> | ||
* Counted result is an integer scalar | * Counted result is an integer scalar | ||
* Generated result is an integer matrix. | * Generated result is an integer matrix. | ||
The count for this function is <apll>!⍠(- | The count for this function is <apll>!⍠(-M) N</apll> where <apll>!⍠(M) N</apll> calculates the [[Variant#Rising_and_Falling_Factorials|'''Rising and Falling Factorial''']]. | ||
For example: | For example: | ||
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3 1 2 | 3 1 2 | ||
3 2 1 | 3 2 1 | ||
⍝ Permutations of length | ⍝ Permutations of length M of N items | ||
⍝ Labeled balls & boxes, ≤1 # Balls per Box | ⍝ Labeled balls & boxes, ≤1 # Balls per Box | ||
!3 | !3 | ||
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</pre> | </pre> | ||
A function to calculate the permutations of <apll> | A function to calculate the permutations of <apll>M</apll> items could be defined as | ||
<pre> | <pre> | ||
perm←{110 1‼⍵ | perm←{110 1‼⍵} | ||
</pre> | </pre> |
Revision as of 17:20, 14 May 2017
This case produces M-Permutations of N items (also called Partial Permutations or Sequences Without Repetition), where when M=N produces the familiar permutations !N. The length of each permutation returned is always M.
- M labeled balls (1), N 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 !⍠(-M) N where !⍠(M) N 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 M of N 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 M items could be defined as
perm←{110 1‼⍵}