Question

有多少个四位整数满足：最左边的数字是奇数，第二个数字是偶数，且四个数字都不同？

How many four-digit whole numbers are there such that the leftmost digit is odd, the second digit is even, and all four digits are different?

$1120$

$1400$

$1800$

$2025$

$2500$

第一位是奇数，有 $5$ 个选择。第二位是偶数，有 $5$ 个选择，而且不会与第一位的奇数重复。

第三位可以是剩下 $8$ 个未用数字中的任意一个，第四位有 $7$ 个选择。总数为 $5 \times 5 \times 8 \times 7 = 1400$ 。

所以正确答案是 B。

The first digit is odd: $5$ choices. The second is even: $5$ choices (none of which repeats the odd first digit).

The third digit is any of the $8$ unused digits, and the fourth is any of the $7$ remaining. In total, $5 \times 5 \times 8 \times 7 = 1400.$

Thus, the correct answer is B .