Hussain2629

Hi all,

can you please help me to find out the number of codes?

mrtwhs

I'll take a stab at this.

3 places for the dot, $\displaystyle \binom{26}{2}$ ways to choose two letters, $\displaystyle \binom{10}{2}$ ways to choose two numbers, and 4! ways to line them up gives all ways allowing a zero at the start. So now we subtract out all those with zero at the start. Again, 3 places for the dot, $\displaystyle \binom{26}{2}$ ways to choose two letters, 9 ways to choose a digit, and 3! ways to line them up. So,

$\displaystyle 3 \cdot \binom{26}{2}\binom{10}{2} \cdot 4! - 3 \cdot \binom{26}{2}\binom{9}{1} \cdot 3! = 100350$