Answer:if we start with 26 letters in the English/Latin alphabet: 26*25*24*23*
and modern Arabic numerals 0–9,
10*9 do the math by multiplying all
Number of dispositions of 4 letters without repetitions of the alphabet (26 letters): 26*25*24*23=358800. 2-digit numbers without repetitions:10*9=90. Number of passwords with 4 letters and 2-digit number without repetitions:
a) if the 2-digit numbers must be in one single position (examples:….XB73DY……..GZ57FE): 358800*90 = 32,292,000.
b) if the 2-digit numbers can be in each of the 5 possible positions (ex.:…73XBDY…X73BDY..XB73DY..XBD73Y..XBDY73…) 5*358800*90=161,460,000.
c) if both digits can be in (C6;2) = 6*5/2= 15 positions (ex.: 7XB3DY…X73BDY…..) 15*358800*90=484,380,000.26*25*24*23*10*9 = 32,292,000
Explanation: