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

There is a total of 26 alphabets. A code word contains 2 English alphabets.$\therefore 2$ alphabets can be filled in ${ }^{26} P _2$
$=\frac{26 !}{(26-2) !}$
$=\frac{26 \times 25 \times 24 !}{24 !}$
$=650 \text { ways. }$
Also, alphabets to be followed by two distinct non-zero digits from 1 to 9 which can be filled in
${ }^9 P_2=\frac{9 !}{(9-2) !}=\frac{9 \times 8 \times 7 !}{7 !}=72$ ways.
∴ Total number of a code words = 650 × 72 = 46800. To find the number of codewords end with an even integer. 2 alphabets can be filled in 650 ways. The digit in the unit’s place should be an even number between 1 to 9, which can be filled in 4 ways. Also, 10’s place can be filled in 8 ways. ∴ Total number of codewords = 650 × 4 × 8 = 20800

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