CombinatorialCase010: Difference between revisions
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This case produces the <apll>L</apll> | This case produces the '''<apll>L</apll> Combinations of <apll>R</apll> items'''. | ||
* <apll>L</apll> unlabeled balls (0), <apll>R</apll> labeled boxes (1), at most one ball per box (0) | * <apll>L</apll> unlabeled balls (0), <apll>R</apll> labeled boxes (1), at most one ball per box (0) |
Revision as of 20:44, 29 April 2017
This case produces the L Combinations of R items.
- L unlabeled balls (0), R labeled boxes (1), at most one ball per box (0)
- Sensitive to ⎕IO
- Count result is an integer scalar
- Generated result is an integer matrix.
The count for this function is L!R.
For example:
If we have 2 unlabeled balls (●●) and 4 labeled boxes (1234) with at most one ball per box, there are 6 (↔ 2!4) ways to meet these criteria:
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and, in general, it’s easy to see that this case solves the familiar problem of L combinations of R items.
The diagram above corresponds to
010 1‼2 4 1 2 1 3 1 4 2 3 2 4 3 4 ⍝ Combinations ⍝ Unlabeled balls, labeled boxes, ≤1 # Balls per Box 3!5 10 010‼3 5 10 010 0‼3 5 10 ⍴010 1‼3 5 10 3 010 1‼3 5 1 2 3 1 2 4 1 2 5 1 3 4 1 3 5 1 4 5 2 3 4 2 3 5 2 4 5 3 4 5
In general, this case is related to that of Multisets (011) and Compositions (012) via the following identities:
010 1‼L R ↔ (011 1‼L,R-L-1)+[⎕IO+1] 0..L-1 011 1‼L R ↔ (010 1‼L,L+R-1)-[⎕IO+1] 0..L-1 010 1‼L R ↔ +\0 ¯1↓012 1‼⍠1 R L+1 012 1‼L R ↔ ¯2-\(010 1‼⍠1 R L-1),L
where ‼⍠1 uses the Variant operator ⍠ to evaluate ‼ in origin 1.