I like this way of computing it better, for several mostly unimportant reasons. :) 

#!/usr/bin/perl -w
use strict;

print countPairPermutations( @ARGV ), $/;

sub countPairPermutations
{
    my( $x, $y )= @_;
    ( $x, $y )= ( $y, $x )
        if  $y < $x;
    my $log= 0;
    for my $i (  1 .. $x  ) {
        $log += log( $y+$i ) - log( $i );
    }
    my $exp= int( $log / log(10) );
    my $mant= exp( $log - log(10)*$exp );
    $mant= sprintf "%.*f", 6 < $exp ? 3 : 8, $mant;
    $mant .=  $exp  ?  "e" . $exp  :  "";
    return $mant
        if  6 < $exp;
    return 0 + $mant;
}

[download]

My code appears to agree with your code (though I didn't count the number of digits output by your code) but they both disagree with your node, reporting that there are 5.807e234 different permutations (not 1.962e230).

Update: Or, using my prototype module: 

#!/usr/bin/perl -w
use strict;
require Math::BigPositiveOkayPrecision;

print countPairPermutations( @ARGV ), $/;

sub countPairPermutations
{
    my( $x, $y )= @_;
    ( $x, $y )= ( $y, $x )
        if  $y < $x;
    my $p= Math::BigPositiveOkayPrecision->new( 1 );
    for my $i (  1 .. $x  ) {
        $p= $p * ( $y+$i ) / $i;
    }
    return $p;
}

[download]

- tye        

In reply to Re^6: One Zero variants_without_repetition (logue) by tye
in thread One Zero variants_without_repetition by thenetfreaker

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