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November 23, 2014

November 23, 2014

Posted by **sweety** on Friday, September 7, 2007 at 1:00pm.

(a) Codes

(b) Codes with distinct letters

(c) Codes with the same letters.

- math -
**drwls**, Friday, September 7, 2007 at 3:09pmThe number of possible different arrangements would be 26 x 26 x 10 = 6760

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