Question

An access code consists of 3 letters of the alphabet followed by 3 digits. (Digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.) How many different access codes are possible?

Answer 1

number of way 3 letters string can be formed from 26 letters of alphabet is 26P3 = 15600

number of way 3 digits string can be formed from 10 digits is 10P3 = 720

So, total possible no of codes is 15600*720 = 11232000 (Assume that repetation of letters and digits not allowed)

{ if repetation are allowed:

1. no of 3 letter string with 3 different no- 15600

2. no of 3 letter string with 2 same no and 1 different no:- 325*6 = 1950
(Lets take 2 letter a,b. possible combination (aab, aba, baa, bba, bab, abb) total 6.
now 2 letter can be choose from 26 letter in 26C2 = 325 ways)

3. no of 3 letter string with 3 same no:- 26

Total letter combination: 15600+1950+26=17576

Similarly total digit combination: 720+(10C2 * 6)+9=999

Possible no of code: 17576*999=17558424