Question

A code consists of 4 digits. first two digits consists of two alphabets and last two digits consists of numbers from 0 to9. how many such codes exist, if repettion of alphabets and the digits is allowed

Answer 1

4209 should be the answer i think.........................not so sure

Answer 2

Answer:

67600 codes exist.

Step-by-step explanation:

Given: A code consist of 4 digits

          First 2 digits contain alphabets and last 2 digits consist of numbers digits from 0 to 9

To find: No of codes when repetition is allowed

We have 26 alphabets

⇒ we 26 ways to select one alphabet at first digit of code and 26 ways for 2nd digit of code.

Total way to select 2 alphabets for 2 digit = 26 × 26 = 676

we have 10 digits or numbers to fill 3rs digit of code and 10 numbers for 4th digit of code.

Total No of way to select last two digit of code = 10 × 10 = 100

Total No way to select a code = 676 × 100 = 67600

Therefore, 67600 codes exist.