AMC10 2009 B

AMC10 2009 B · Q11

AMC10 2009 B · Q11. It mainly tests Basic counting (rules of product/sum), Permutations.

How many 7 digit palindromes (numbers that read the same backward as forward) can be formed using the digits 2, 2, 3, 3, 5, 5, 5?

使用数字 2, 2, 3, 3, 5, 5, 5 可以形成多少个7位回文数（正读反读都相同的数字）？

(A) 6 6

(B) 12 12

(D) 36 36

(E) 48 48

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Because the digit 5 appears three times, 5 must be the middle digit of any such palindrome. In the first three digits each of 2, 3, and 5 must appear once and the order in which they appear determines the last three digits. Since there are $3! = 6$ ways to order three distinct digits the number of palindromes is 6.

答案（A）：因为数字 5 出现了三次，所以在任何这样的回文数中，5 必须是中间一位。在前三位数字中，2、3、5 必须各出现一次，它们出现的顺序决定了后三位数字。由于三个不同数字的排列方式有 $3! = 6$ 种，因此回文数的个数是 6。

Topics