AMC8 2020

AMC8 2020 · Q7

AMC8 2020 · Q7. It mainly tests Combinations.

How many integers between $2020$ and $2400$ have four distinct digits arranged in increasing order? (For example, $2347$ is one integer.).

2020 到 2400 之间有多少个四位不同数字且按升序排列的整数？（例如，2347 就是一个这样的整数。）

(A) \text{9} \text{9}

(B) \text{10} \text{10}

(D) \text{21} \text{21}

(E) \text{28} \text{28}

Answer

Correct choice: (C)

正确答案：（C）

Solution

Firstly, we can observe that the second digit of such a number cannot be $1$ or $2$ because the digits must be distinct and in increasing order. The second digit also cannot be $4$ as the number must be less than $2400$, so the second digit must be $3$. It remains to choose the latter two digits, which must be $2$ distinct digits from $\left\{4,5,6,7,8,9\right\}$. That can be done in $\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15$ ways; there is then only $1$ way to order the digits, namely in increasing order. This means the answer is $\boxed{\textbf{(C) }15}$.

首先，我们观察到此类数字的第二位数字不能是 $1$ 或 $2$，因为数字必须不同且按升序排列。第二位数字也不能是 $4$，因为数字必须小于 $2400$，所以第二位必须是 $3$。剩下选择后两位数字，必须从 $\left\{4,5,6,7,8,9\right\}$ 中选择 $2$ 个不同数字。有 $\binom{6}{2} = \frac{6 \cdot 5}{2 \cdot 1} = 15$ 种方式；然后数字只有 $1$ 种排序方式，即升序。因此答案是 $\boxed{\textbf{(C) }15}$。

Topics

CP.003 Combinations