user
You need a random arrangement of a deck of cards, you are sick of lame ways of doing this. This task is a super-cool way of doing this using factorial base numbers.
The first 25 factorial base numbers in increasing order are: 0.0.0, 0.0.1, 0.1.0, 0.1.1, 0.2.0, 0.2.1, 1.0.0, 1.0.1, 1.1.0, 1.1.1,1.2.0, 1.2.1, 2.0.0, 2.0.1, 2.1.0, 2.1.1, 2.2.0, 2.2.1, 3.0.0, 3.0.1, 3.1.0, 3.1.1, 3.2.0, 3.2.1, 1.0.0.0
Observe that the least significant digit is base 2 the next base 3, in general an n-digit factorial base number has digits n..1 in base n+1..2.

I want to produce a 1 to 1 mapping between these numbers and permutations:-

```
       0.0.0 -> 0123
       0.0.1 -> 0132
       0.1.0 -> 0213
       0.1.1 -> 0231
       0.2.0 -> 0312
       0.2.1 -> 0321
       1.0.0 -> 1023
       1.0.1 -> 1032
       1.1.0 -> 1203
       1.1.1 -> 1230
       1.2.0 -> 1302
       1.2.1 -> 1320
       2.0.0 -> 2013
       2.0.1 -> 2031
       2.1.0 -> 2103
       2.1.1 -> 2130
       2.2.0 -> 2301
       2.2.1 -> 2310
       3.0.0 -> 3012
       3.0.1 -> 3021
       3.1.0 -> 3102
       3.1.1 -> 3120
       3.2.0 -> 3201
       3.2.1 -> 3210

```

The following psudo-code will do this:
Starting with m=0 and Ω, an array of elements to be permutated, for each digit g starting with the most significant digit in the factorial base number.

If g is greater than zero, rotate the elements from m to m+g in Ω (see example)
Increment m and repeat the first step using the next most significant digit until the factorial base number is exhausted.
For example: using the factorial base number 2.0.1 and Ω = 0 1 2 3 where place 0 in both is the most significant (left-most) digit/element.

Step 1: m=0 g=2; Rotate places 0 through 2. 0 1 2 3 becomes 2 0 1 3
Step 2: m=1 g=0; No action.
Step 3: m=2 g=1; Rotate places 2 through 3. 2 0 1 3 becomes 2 0 3 1

Let me work 2.0.1 and 0123

```
    step 1 n=0 g=2 Ω=2013
    step 2 n=1 g=0 so no action
    step 3 n=2 g=1 Ω=2031

```

The task:

```
 First use your function to recreate the above table.
 Secondly use your function to generate all permutaions of 11 digits, perhaps count them don't display them, compare this method with
    methods in rc's permutations task.
 Thirdly here following are two ramdom 51 digit factorial base numbers I prepared earlier:
   39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0
   51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1
   use your function to crate the corresponding permutation of the following shoe of cards:
      A♠K♠Q♠J♠10♠9♠8♠7♠6♠5♠4♠3♠2♠A♥K♥Q♥J♥10♥9♥8♥7♥6♥5♥4♥3♥2♥A♦K♦Q♦J♦10♦9♦8♦7♦6♦5♦4♦3♦2♦A♣K♣Q♣J♣10♣9♣8♣7♣6♣5♣4♣3♣2♣
 Finally create your own 51 digit factorial base number and produce the corresponding permutation of the above shoe

```
assistant
(formerly Perl 6)

**Works with**: Rakudo version 2018.11

Using my interpretation of the task instructions as shown on the discussion page.

```
sub postfix:<!> (Int $n) { (flat 1, [\*] 1..*)[$n] }

multi base (Int $n is copy, 'F', $length? is copy) {
    constant @fact = [\*] 1 .. *;
    my $i = $length // @fact.first: * > $n, :k;
    my $f;
    [ @fact[^$i].reverse.map: { ($n, $f) = $n.polymod($_); $f } ]
}

sub fpermute (@a is copy, *@f) { (^@f).map: { @a[$_ .. $_ + @f[$_]].=rotate(-1) }; @a }

put "Part 1: Generate table";
put $_.&base('F', 3).join('.') ~ ' -> ' ~ [0,1,2,3].&fpermute($_.&base('F', 3)).join for ^24;

put "\nPart 2: Compare 11! to 11! " ~ '¯\_(ツ)_/¯';
# This is kind of a weird request. Since we don't actually need to _generate_
# the permutations, only _count_ them: compare count of 11! vs count of 11!
put "11! === 11! : {11! === 11!}";

put "\nPart 3: Generate the given task shuffles";
my \Ω = <A♠ K♠ Q♠ J♠ 10♠ 9♠ 8♠ 7♠ 6♠ 5♠ 4♠ 3♠ 2♠ A♥ K♥ Q♥ J♥ 10♥ 9♥ 8♥ 7♥ 6♥ 5♥ 4♥ 3♥ 2♥
         A♦ K♦ Q♦ J♦ 10♦ 9♦ 8♦ 7♦ 6♦ 5♦ 4♦ 3♦ 2♦ A♣ K♣ Q♣ J♣ 10♣ 9♣ 8♣ 7♣ 6♣ 5♣ 4♣ 3♣ 2♣
>;

my @books = <
    39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0
    51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1
>;

put "Original deck:";
put Ω.join;

put "\n$_\n" ~ Ω[(^Ω).&fpermute($_.split: '.')].join for @books;

put "\nPart 4: Generate a random shuffle";
my @shoe = (+Ω … 2).map: { (^$_).pick };
put @shoe.join('.');
put Ω[(^Ω).&fpermute(@shoe)].join;

put "\nSeems to me it would be easier to just say: Ω.pick(*).join";
put Ω.pick(*).join;

```

Output:

```
Part 1: Generate table
0.0.0 -> 0123
0.0.1 -> 0132
0.1.0 -> 0213
0.1.1 -> 0231
0.2.0 -> 0312
0.2.1 -> 0321
1.0.0 -> 1023
1.0.1 -> 1032
1.1.0 -> 1203
1.1.1 -> 1230
1.2.0 -> 1302
1.2.1 -> 1320
2.0.0 -> 2013
2.0.1 -> 2031
2.1.0 -> 2103
2.1.1 -> 2130
2.2.0 -> 2301
2.2.1 -> 2310
3.0.0 -> 3012
3.0.1 -> 3021
3.1.0 -> 3102
3.1.1 -> 3120
3.2.0 -> 3201
3.2.1 -> 3210

Part 2: Compare 11! to 11! ¯\_(ツ)_/¯
11! === 11! : True

Part 3: Generate the given task shuffles
Original deck:
A♠K♠Q♠J♠10♠9♠8♠7♠6♠5♠4♠3♠2♠A♥K♥Q♥J♥10♥9♥8♥7♥6♥5♥4♥3♥2♥A♦K♦Q♦J♦10♦9♦8♦7♦6♦5♦4♦3♦2♦A♣K♣Q♣J♣10♣9♣8♣7♣6♣5♣4♣3♣2♣

39.49.7.47.29.30.2.12.10.3.29.37.33.17.12.31.29.34.17.25.2.4.25.4.1.14.20.6.21.18.1.1.1.4.0.5.15.12.4.3.10.10.9.1.6.5.5.3.0.0.0
A♣3♣7♠4♣10♦8♦Q♠K♥2♠10♠4♦7♣J♣5♥10♥10♣K♣2♣3♥5♦J♠6♠Q♣5♠K♠A♦3♦Q♥8♣6♦9♠8♠4♠9♥A♠6♥5♣2♦7♥8♥9♣6♣7♦A♥J♦Q♦9♦2♥3♠J♥4♥K♦

51.48.16.22.3.0.19.34.29.1.36.30.12.32.12.29.30.26.14.21.8.12.1.3.10.4.7.17.6.21.8.12.15.15.13.15.7.3.12.11.9.5.5.6.6.3.4.0.3.2.1
2♣5♣J♥4♥J♠A♠5♥A♣6♦Q♠9♣3♦Q♥J♣10♥K♣10♣5♦7♥10♦3♠8♥10♠7♠6♥5♠K♥4♦A♥4♣2♥9♦Q♣8♣7♦6♣3♥6♠7♣2♦J♦9♥A♦Q♦8♦4♠K♦K♠3♣2♠8♠9♠

Part 4: Generate a random shuffle
47.9.46.16.28.8.36.27.29.1.9.27.1.16.21.22.28.34.30.8.19.27.18.22.3.25.15.20.12.14.8.9.11.1.4.0.3.5.4.2.2.10.8.1.6.1.2.4.1.2.1
6♣5♠5♣10♥10♦6♠K♣9♦6♦K♠2♠5♦Q♠5♥Q♦8♦J♣2♣8♣A♥K♦9♣A♦2♦9♠4♣3♥A♣7♥2♥Q♥9♥4♥J♠4♠A♠3♠8♥J♥7♠K♥3♣10♣8♠Q♣6♥7♦7♣J♦3♦4♦10♠

Seems to me it would be easier to just say: Ω.pick(*).join
5♦3♠8♦10♦2♥7♠7♦Q♦A♠5♣8♣Q♠4♠2♦K♦5♠Q♥7♣10♠2♠K♠J♣9♣3♣4♥3♥4♦3♦Q♣2♣4♣J♦9♠A♣J♠10♣6♣9♦6♠10♥6♥9♥J♥7♥K♥A♦8♠A♥5♥8♥K♣6♦
```