2009 AIME I Problems/Problem 10
Revision as of 14:01, 20 March 2009 by Thirdgrademustang (talk | contribs) (New page: Since after each planet, only members of another planet can follow, we simply count the lengths of the blocks adding up to ten. We consider a few different cases: 1. Block of 5 - There is ...)
Since after each planet, only members of another planet can follow, we simply count the lengths of the blocks adding up to ten. We consider a few different cases: 1. Block of 5 - There is only one way to arrange this so 1^3=1
2. 5 Blocks of 1 - There is also only one way to arrange this so we get 1^3=1
3. 2 Blocks - There are two cases: 4+1 and 3+2. Each of these can be arranged two ways so we get (2+2)^3=64
4. 3 Blocks - There are also two cases: 3+1+1 and 2+2+1.Each of these can be arranged 3 ways giving us (3+3)^3=216
5. 4 Blocks - There is only one case: 2+1+1+1. This can be arranged 4 ways giving us 4^3=64
Combining all these cases, we get 1+1+64+64+216=346
