Mock AIME 1 2010 Problems/Problem 8

Revision as of 15:32, 11 August 2017 by Greenturtle3141 (talk | contribs) (Previous solution only accounted for order?)

Solution

Once the pieces are placed, there will be $20-3(3)-2(2)-1=6$ blank spaces. So, we are simply ordering $3$ triominoes, $2$ dominoes, $1$ square, and $6$ blank spaces. That's just $\frac{12!}{3!2!6!} = 55440$, with last three digits $\boxed{440}$.