# Difference between revisions of "2009 AMC 10B Problems/Problem 11"

## Problem

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

$\text{(A) } 6 \qquad \text{(B) } 12 \qquad \text{(C) } 24 \qquad \text{(D) } 36 \qquad \text{(E) } 48$

## Solution

A seven-digit palindrome is a number of the form $\overline{abcdcba}$. Clearly, $d$ must be $5$, as we have an odd number of fives. We are then left with $\{a,b,c\} = \{2,3,5\}$. Each of the $\boxed{(A)6}$ permutations of the set $\{2,3,5\}$ will give us one palindrome.