Question

A code word is formed by two different English letters followed by two non-zero distinct digits. Find the number of such code words. Also, find the number of such code words that end with an even digit.

Answer 1

There are 26 alphabets and 9 non-zero digits from 1 to 9.

For a code, 2 distinct alphabets can be arranged in ²⁶P₂ ways and 2 digits can be arranged in ⁹P₂ ways.

Hence, the total number of codewords available

= ²⁶P₂ × ⁹P₂

= `(26!)/(24!) xx (9!)/(7!)`

= `(26 xx 25 xx 24!)/(24!) xx (9 xx 8 xx 7!)/(7!)`

= (26 × 25) × (9 × 8)

= 650 × 72

= 46800

For a code, 2 alphabets can be arranged in ²⁶P₂ ways, and of the, 2 distinct digits the last place can be filled by either 2, 4, 6, 8, i.e., 4 ways, while the first place of the digit can be filled from the remaining 8 digits in 8 ways.

Hence, the total number of code words that end with an even integer

= ²⁶P₂ × 4 × 8

= 26 × 25 × 4 × 8

= 20800