SOLUTION: How many 4 - digit even numbers can be formed from the digits 0 to 9 if each digit is to be used only once in each number ?

The fact that the number is an EVEN number means that the last digit is one of 5 even digits 0, 2, 4, 6, or 8.


In my solution, I will consider two cases separately:

        case (a):  the last digit is 0 (zero),

    and

        case (b): the last digit is any of the remaining 4 even digits 2, 4, 6 or 8.

    Then the first (most-left) digit is any of 9 remaining digits;

         the second digit is any of remaining 8 digits;

         the third digit is any of remaining 7 digits.


    So, the total number of possible options is  9*8*7 = 504 in this case.

    Then the first (most-left) digit is any of 8 remaining digits (keep in mind that the leading digit CAN NOT be 0 (!));

         the second digit is any of 8 remaining digits (zero is ALLOWED in this position);

         the third digit  is any of 7 remaining digits (zero is ALLOWED in this position).


    So, the total number of possible options is  8*8*7 = 448 in this case.