how many are there

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Hi I was wondering does anyone know how many of each in a 10 million prefix. 
 2 digit radars
 3 digit radars
4 digit radars
repeater note 3 cycle
repeater note 3-cycle that is not a radar

thanks for your help

cheers

polarbear

I counted these up in my rotators article in the June 2004 issue of the Canadian Paper Money Society Newsletter:

... there are only 630 (10x9x7) two-digit radar notes per 10,000,000 notes printed. You have 10 choices for the first digit, 9 choices for the second digit and there are 7 distinct ways to arrange those digits. Without going into a mathematical derivation, assume that there are also 5,040 4-digit radars (10x9x8x7) and 4,320 3-digit radars (10x9x8x6) and 9 solid radars for a total of 9,999 radars per 10,000,000 notes. (Again, BA Banknote will reduce this number to about 9,990 radars because it destroys the entire sheet containing 0000000 and 9 other radars).

You can find my article here: http://www.bwjm.ca/pub_rotators.asp

[edit]I have determined that there was some degree of error in my numbers for the repeaters that I originally posted. This has been removed and I have reposted a better (and hopefully correct) solution below.[/edit]

As for repeaters, this is a bit tricky. Try to stay with me here...

NOTE: I only refer to n-cycle repeaters, not n-digit repeaters. The cycles are far easier to count than the digits. 1121121 is a 3-cycle 2-digit repeater. 1212121 is a 2-cycle 2-digit repeater.

a) 1-cycle repeaters (AAAAAAA)
These are the same as solids, or 1-digit radars. There are 10, including the normally-destroyed 0000000 note.
You have 10 choices for the first digit, and that digit is the same for the rest of the serial number. Pretty easy.
These are all solid-numbered notes and 1-digit radars.

b) 2-cycle repeaters (ABABABA)
You have 10 choices for the first digit, 9 choices for the second digit, and then the pattern repeats. 10x9 = 90. Still pretty easy.
These are all 2-digit radars, but this is not all possible 2-digit radars.

c) 3-cycle repeaters (ABCABCA)
You thought I'd do 10x9x8 here, didn't you? Not quite. Here we're counting all the numbers such that we get a pattern of ABCABCA except for the case where A=B=C (this is then just AAAAAAA which we already counted). So, we've got 10 choices for the first digit, 10 choices for the second, and 10 for the third, with the pattern then repeating. We need to subtract the 1-cycle repeaters we've already counted. 10x10x10 - 10 = 990. Did I lose anyone?
Some of these will be 2-digit radars if B=C. (ie: 1221221)

d) 4-cycle repeaters (ABCDABC)
Here we've got something similar to the 3-cycle repeaters: 10 choices for each of the first four digits. That's a grand total of 10,000 notes. However, we're double counting some. If A=C and B=D then we've got ABABABA, which is already counted, so we subtract them. If A=B=C=D, that's AAAAAAA which also is already counted, so we subtract them.
You're probably asking "what if A=D?" (or something similar). Well, that's ABCAABC, which is still merely a 4-cycle repeater, and we haven't already counted that. 10x10x10x10 - 90 - 10 = 9,900.
Some of these will be 3-digit radars if A=C. (ie: 0102010)

The Bottom Line:
1-Cycle Repeaters: 10
2-Cycle Repeaters: 90
3-Cycle Repeaters: 990
4-Cycle Repeaters: 9,900

These counts were proven by a script I wrote to exhaustively list all repeaters from 0000000 to 9999999.