How many even 3-digit numbers can be formed from digits 0 1 2 4 5 7 9 if each digit can be used only once?

Answer 1

To solve this question, two cases must be considered:

Case 1: The three-digit even number ends with 0.

If the three-digit number ends with 0, then the number must be even. After the digit 0 is used, six digits remain. After any one of these six digits is chosen to be the first digit, five digits remain. To complete the number, any of the five remaining digits could be chosen to be the second digit of the number.

___6___ * ___5___ * ___1___ = 30 three-digit even numbers ending in 0.

1st digit-------2nd digit----3rddigit (0)

Case 2: The three-digit even number does not end with 0.

If the three-digit number does not end with 0, it can end with 2 or 4. Therefore, there are two possibilities for the third digit. The first digit could be any of the remaining digits, EXCEPT 0: if 0 were the first digit, it would not be a three-digit number. Therefore, there would be five possible digits for the first digit. 0 could be used as the second digit, along with the four remaining digits that were not previously used.

___5___ * ___5___ * ___2___ = 50 three-digit even numbers not ending in 0.

1st digit-------2nd digit-----3rddigit (0)

The total number of three-digit even numbers is 80, since 30 three-digit even numbers ending in 0 plus 50 three-digit even numbers not ending in 0 equals 80.