6. How many numbers of the form $\overline{2 a b c d 3}$ are there such that all digits are distinct, and the number $\overline{a b c d}$ is a four-digit multiple of 5?
6. How many numbers of the form $\overline{2 a b c d 3}$ have all digits mutually distinct, and the number $\overline{a b c d}$ is a four-digit multiple of 5?

## Solution.

The number $\overline{a b c d}$ is a multiple of 5, so $d \in\{0,5\}$.

The number $\overline{a b c d}$ is a four-digit number with all digits mutually distinct, so if

$$
d=0
$$

the digits $a, b, c$ belong to the set $\{1,4,5,6,7,8,9\}$

1 POINT

The digit $a$ can be chosen in 7 ways, the digit $b$ in 6 ways, and the digit $c$ in 5 ways.

There are $7 \cdot 6 \cdot 5=210$ numbers of the form $\overline{2 a b c 03}$.

3 POINTS

If

$$
d=5
$$

the digits $a, b, c$ belong to the set $\{0,1,4,6,7,8,9\}, a \neq 0$.

1 POINT

The digit $a$ can be chosen in 6 ways, the digit $b$ in 6 ways, and the digit $c$ in 5 ways.

There are $6 \cdot 6 \cdot 5=180$ numbers of the form $\overline{2 a b c 53}$.

3 POINTS

There are $210+180=390$ numbers of the form $\overline{2 a b c d 3}$ with all digits mutually distinct.
1 POINT
TOTAL 10 POINTS