Question

A code word should consist of two English distinct capital alphabets followed by two distinct digits from $1$ to $9$ e.g. $MH23$ is a code word.
(a) How many such code words are available ?
(b) How many of them end with an even integers?

Answer 1

(a) First letter of the word could be any one the 26.

English alphabets.

Second letter of the word could be any one of the 25.

English Alphabets other than the first letter.

First digit could be any of 1 to 9 which can be chosen in

9 different ways

Second digit could be any of 1 to 9 other than the first digit, so second digit can be chosen in 8 different ways.

So, either selection of first letter, second letter,

First digit and second digit can happen in

$=26\times 25\times 9\times 8$ ways

$=46,800$

(b) To end with an even digit, let us fix the second digit

first could be any of 2,4,6 or 8, once the second digit chosen, first digit can be chosen in 8 different way and the first 2 letter can be chosen in $26\times 25$

ways.

So, the number of different ways of selecting in which last digit is even are = $\mathrm{26.25.8.4}=20,800$