AMC 8 2004 Problem 2
AMC8 2004 年第 2 题
排列与组合★★☆☆☆
How many different four-digit numbers can be formed by rearranging the four digits in 2004 ?
将 2004 的四个数字重新排列可以形成多少个不同的四位数?
- A.4
- B.6
- C.16
- D.24
- E.81
正确答案 · Correct Answer
B
解析 · Solution
Solution 1
We can solve this problem easily, just by calculating how many choices there are for each of the four digits. First off, we know there are only 2 choices for the first digit, because 0 isn't a valid choice, or the number would a 3-digit number, which is not what we want. We have 3 choices for the second digit, since we already used up one of the digits, and 2 choices for the third, and finally just 1 choices for the fourth and final one. 2*3*2*1 is 12, but there are 2 zeros that have been counted as different numbers, so divide by 2 to get 6.
Note that the four-digit number must start with either a 2 or a 4. The four-digit numbers that start with 2 are 2400, 2040, and 2004. The four-digit numbers that start with 4 are 4200, 4020, and 4002 which gives us a total of 6.
In order for the resulting numbers to have four digits, they cannot start with 0. Therefore, both zeroes must be in the last three places. There are ways to choose which two of the last three places are zeroes. Then there are ways to arrange the 2 and the 4 in the remaining two places, giving us a total of .
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