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:
-3(c-10) + -4= 2
-3c+ 30+ -4= 2
-3c + 26= 2
-3c= -24
C= 8
✌HEYA!!!
HERE IS YOUR ANSWER=
THERE ARE ONLY TWO CASES WHEN COMPUTER PICKED A 1 AND A 2 OUT OF 25..
CASES= (1,2);(2,1)
P (E)=2/25
O.O8
WHICH IS OPTION D.
HOPE IT HELPS YOU '_'
I’m not 100% sure this is correct, because I’ve never done these before, but try 12, this is the answer I got from a reliable app that I usually use