A code word is to consist of two alphabets followed by two distinct numbers between 1 and 9 for example, CA23 is a code. How many such code words are there?

Hint: In the given question we have to use the concept of permutation and combination. In this question first we have to choose 2 alphabets from 26 English alphabets. When we select 2 alphabets, then we move to select 2 numbers from 0 to 9. After completing the selection process we totally select 4 letters and different code can be formed by randomizing these numbers and letters.

Complete step by step solution: As in this following question, firstly we can notice that first two alphabets repetition of letter is allowed. So, the first two places of the alphabet can be repeated. But for the digit value cases will be different.
Step 1: We are going to select 2 letters from 26 alphabets where repetition of letters is allowed. For the first letter there are 26 choices and for the second letter also there are 26 choices. Hence, it is compulsory to multiply both numbers of selection ways.
$\begin{array}{l}
{ = ^{26}}{C_1}{ \times ^{26}}{C_1}\\
 = 26 \times 26\\
 = 676
\end{array}$
Step 2: Now, we move to the 2nd part of the question, which is selection of 2 distinct numbers from 1 to 9.
$ = \left( {^9{C_1}{ \times ^8}{C_1}} \right)$ (Using multiplication theorem)
$\begin{array}{l}
 = 9 \times 8\\
 = 72
\end{array}$
Step 3: Now we are going to form four (2 letters + 2 digits) character code. To form four character code, multiplying both step 1 and step 2 are compulsory because of the formation of the code:
∴ Total numbers of code form are
$\begin{array}{l}
 = 26 \times 26 \times 72\\
 = 676 \times 72\\
 = 48,672
\end{array}$
Hence, the total number of code words possible is 48,672.

Note: In this type of question, a student must be careful about whether repetition of a letter is allowed or not. When there are two such conditions, then occurrences of both are compulsory where both are going to be multiplied.