AMC8 1995

AMC8 1995 · Q23

AMC8 1995 · Q23. It mainly tests Basic counting (rules of product/sum), Casework.

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?

有多少个四位整数，使得最左边数字是奇数，第二个数字是偶数，并且所有四个数字互不相同？

(A) 1120 1120

(B) 1400 1400

(D) 2025 2025

(E) 2500 2500

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): There are 5 odd and 5 even digits that can be used for the two leftmost digits in the number. Once an odd and even digit have been selected and since all four digits are different, there are 8 choices remaining for the third digit, and then 7 choices for the fourth digit. Thus there are $5\times5\times8\times7=1400$ such whole numbers.

答案（B）：可用于该数最左边两位的奇数有5个、偶数也有5个。选定一个奇数和一个偶数后，由于四个数字都不同，第三位还剩8种选择，第四位有7种选择。因此这样的整数共有$5\times5\times8\times7=1400$个。

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