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Posted by **Anonymous** on Thursday, January 8, 2009 at 2:11pm.

- Math -
**bobpursley**, Thursday, January 8, 2009 at 3:37pmC=27!/25! * 10!/8! * 4!/2!2!

The last term represents the number of permuations of two letters, two numbers can be done (six ways).

- Math -
**Reiny**, Thursday, January 8, 2009 at 4:13pmmy approach:

first lets choose 2 letters and 2 numbers,

that is C(26,2)*C(10,2)

= 325*45

= 14625

let's consider one of these, say, DK45

now that could be arranged in 4! ways

so the total number of passwords is

14625*4! = 351000

bobpursley, when did they introduce that 27th letter?

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