Hi. Just two cents...

The frequencies of the individual digits will be certain patterns
limited to some combination ...

These are the "integer Partitions" of the number 4. (the numbers that
sum to 4).
The number 4 has 5 as seen here. A number like 10 has 42, etc.

{{4}, {3, 1}, {2, 2}, {2, 1, 1}, {1, 1, 1, 1}}

a unique pattern we can use to get the perms.

Here's the general equation. We skip 1! since it's one. (I used it in
the last example thou)

7777 1
1777 4
0044 6
2477 12
1234 24


4!/(4!)
1

4!/(3!)
4

4!/(2! 2!)
6

4!/(2!)
12

4!/(1! 1! 1! 1!)
24

= = = = = = = =
HTH :)
Dana DeLouis


On 11/29/09 2:09 PM, T. Valko wrote:
Think of the 4 digit number as a 4 card poker hand. These would be the best
possible hands:

1111 = 4 of a kind
1112 = 3 of a kind
1122 = 2 pairs
1123 = 1 pair
1234 = high card

The frequencies of the individual digits will be certain patterns limited to
some combination of:

1111 = 4;0;0;0;0
1112 = 3;0;0;1;0
1122 = 2;0;2;0;0
1123 = 2;0;1;1;0
1234 = 1;1;1;1;0

So, we only need to look for the two highest numbers of the frequencies to
come up with a unique pattern we can use to get the perms.

LARGE(...,1)*10 + LARGE(...,2)*1

1111 = 4;0;0;0;0 = (4*10) + (0*1) = 40
1112 = 3;0;0;1;0 = (3*10) + (1*1) = 31
1122 = 2;0;2;0;0 = (2*10) + (2*1) = 22
1123 = 2;0;1;1;0 = (2*10) + (1*1) = 21
1234 = 1;1;1;1;0 = (1*10) + (1*1) = 11

Then it's just a simple lookup:

..........J..........K
1......11.........24 = high card
2......21.........12 = 1 pair
3......22.........6 = 2 pairs
4......31.........4 = 3 of a kind
5......40.........1 = 4 of a kind