CountTheOpposite MultiplicationPrinciple AdditionPrinciple Basic

Problem - 4771

How many even $4$- digit integers are there whose digits are distinct?

Answer
$2296$

We have two restrictions here: the first digit cannot be $0$ and the last digit must be even.

**Solution 1**

We first choose the first digit, then the last digit, afterwards the second and the third digit. However, because $0$ is an even digit itself but cannot be placed as the first, we need to do a casework:

last digit is $0$:

- Last digit: $1$ choice
- First digit: $9$ choices
- Second digit: $8$ choices
- Third digit: $7$ choices

last digit is not $0$:

- Last digit: $4$ choices: $2$, $4$, $6$, and $8$
- First digit: $8$ choices: canot be zeo
- Second digit: $8$ choices
- Third digit: $7$ choices

Hence the answerr is $1\times 9 \times 8\times 7 + 4\times 8\times 8\times 7 = \boxed{2296}$.

**Solutions 2**

Alternatively, we can also employ the count the opposite technique because counting odd number is easier.

- Total $4$-digit numbers: $9\times 9\times 8\times 7=4536$ (steps: first digit, second digit, third digit, last digit)
- Total odd numbers: $5\times 8\times 8\times 7 =2240$ (steps: last digit, first, second, third)

Therefore, the number of odd integers: $4536-2240 =2296$.