7a+3b=27 6a+3b=24 Solve in simultaneous A= B=
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The number of permutations of the 25 letters taken 2 at a time (with repetitions) is:
The number of permutations of the 9 digits taken 4 at a time (with repetitions) is:
Each permutation of letters can be taken with each permutation of digits, therefore the total number of possible passwords is:
The number of permutations of the 9 digits taken 4 at a time (with repetitions) is:
Each permutation of letters can be taken with each permutation of digits, therefore the total number of possible passwords is: