Question

2) A passcode consists of three letters (A-Z, a total of 26) followed the three digits (0-9). Repetitions are allowed. How many such passcodes are possible? (You don't need to simplify your answer)

Answer 1

when number of distinct alphabets=1 then three numbers can be chosen in 26 ways.

when number of distinct alphabets=2 then three numbers can be chosen in $\binom{26}{2}$ ways and they(2 similar, 1different) can be arranged in 3 ways.

when number of distinct alphabets=3 then three numbers can be chosen in $\binom{26}{3}$ ways and they(all 3 are different) can be arranged in 3!=6 ways.

Therefore total number of arrangements of 3 alphabets=

26 26+ 26 2 *3+ *6 3

Similarly,

when number of distinct digit=1 then three numbers can be chosen in 10 ways.

when number of distinct digits=2 then three numbers can be chosen in $\binom{10}{2}$ ways and they(2 similar, 1different) can be arranged in 3 ways.

when number of distinct digits=3 then three numbers can be chosen in $\binom{10}{3}$ ways and they(all 3 are different) can be arranged in 3!=6 ways.

Therefore total number of arrangements of 3 digits=$10+\binom{10}{2}*3+\binom{10}{3}*6$

Hence total number of passcodes=$\{26+\binom{26}{2}*3+\binom{26}{3}*6\}*\{10+\binom{10}{2}*3+\binom{10}{3}*6\}$