2009
Problem - 1570
How many $7$-digit palindromes (numbers that read the same backward as forward) can be formed using the digits $2$, $2$, $3$, $3$, $5$, $5$, $5$?
Clearly, one $5$ must be in the middle. Then there are three places before this middle $5$ which must be occupied by $2$, $3$, and $5$ in some order. Therefore, the answer is $3!=\boxed{6}$.