CountPermutations ( string )

To determine the number of permutations for a string of characters

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Maarten Witberg - Show more from this author
http://www.maatwerkdbs.nl

Sample input:

CountPermutations ( "12341" )

Sample output:

Function definition: (Copy & paste into FileMaker's Edit Custom Function window)

//CountPermutations ( string ) 
//cf by maarten witberg 2008
//to determine the number of permutations of a string of characters

Let (
[ 
sub = CountPermutationsSub ( string ) //call to sub function (see below)

];

Evaluate 
( 
   Left  
         ( 
           "((((((((((((((((((((((((((((((((((((" ; PatternCount (  sub ; ")" ) // number of parentheses depends on how many characters are involved e.g. a-z, 0-9
         ) // end left
     &
   Factorial ( Length ( string) ) & sub
) // end eval

) // end let

/*
Define this sub function first:
//CountPermutationsSub ( string ) = 
//cf by maarten witberg 2008
//called by CountPermutations ( string )
Let (
[ 
dim = Left ( string ; 1 ) ;
ptc = Factorial ( PatternCount ( string ; dim ) ) ; 
list=   
          Case ( 
                     Length (string ) > 1 ; 
                     CountPermutationsSub ( GetAsNumber ( Substitute ( string ; dim ; "") ) ) 
                   ) 
           & "/" &  ptc & ")"

];
list
)
*/

needs sub function (embedded here for efficiency). algorithm builds a formula along the lines of

((....( stringlength! / numberofduplicate1! ) / numberofduplicate2!)/.........)/numberofduplicateN!)
then evaluates it.

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CountPermutations ( string )

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