Difference between revisions of "Mock AIME 1 2010 Problems/Problem 8"

(Created page with "==Solution== This is the same thing as arranging 1 A, 2 Bs, and 3 Cs to form a word. Our answer is <math>\frac{6!}{3!2!1!}=\boxed{60}</math>.")
 
(Previous solution only accounted for order?)
 
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==Solution==
 
==Solution==
  
This is the same thing as arranging 1 A, 2 Bs, and 3 Cs to form a word. Our answer is <math>\frac{6!}{3!2!1!}=\boxed{60}</math>.
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Once the pieces are placed, there will be <math>20-3(3)-2(2)-1=6</math> blank spaces. So, we are simply ordering <math>3</math> triominoes, <math>2</math> dominoes, <math>1</math> square, and <math>6</math> blank spaces. That's just <math>\frac{12!}{3!2!6!} = 55440</math>, with last three digits <math>\boxed{440}</math>.

Latest revision as of 16:32, 11 August 2017

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 $\boxed{440}$.