Build a hash that relates the permutation to its position, and use that to decide which count to increment:

#! perl -slw
use strict;
use Data::Dump qw[ pp ];
use Algorithm::Combinatorics qw[ variations_with_repetition ];

our $N //= 4;

my @perms = map join( '', @$_ ), variations_with_repetition( [ 1 .. $N
+ ], $N );

my %lookup;
$lookup{ $perms[ $_ ] } = $_ for 0 .. $#perms;

my @counts = (0) x @perms;

for( 1 .. 1e3 ) {
    my $observed = join '', map 1+int( rand $N ), 1 .. $N; ## random o
+bservation
    ++$counts[ $lookup{ $observed } ];
}

printf "%${N}s : %${N}d\n", $perms[ $_ ], $counts[ $_ ]
    for 0 .. $#perms;
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A run:

C:\test>junk -N=3
111 :  41
112 :  37
113 :  35
121 :  27
122 :  32
123 :  31
131 :  43
132 :  34
133 :  45
211 :  43
212 :  35
213 :  40
221 :  47
222 :  38
223 :  32
231 :  46
232 :  44
233 :  30
311 :  33
312 :  37
313 :  30
321 :  30
322 :  49
323 :  35
331 :  42
332 :  32
333 :  32
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