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 integer?

Answer 1

Answer:

(a) 46800

(b) 20800

Step-by-step explanation:

Hi,

(a)

First letter of the word could be any one of 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, entire selection of First letter , Second Letter, First

digit and second digit can happen in

26*25*9*8 ways = 46800

(b)

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

first , that could be any of 2, 4 , 6 or 8 , Once the second

digit is chosen, the first digit can be chosen in 8 different

ways, and the first 2 letter can be chosen in 26*25 distinct

ways,

So, total number of different ways of selecting

in which last digit is even are 26*25*8*4 = 20800

Hope, it helps !