AIME II 2009 #6

[stupidityismygam]

Let [unparseable or potentially dangerous latex formula] be the number of five-element subsets that can be chosen from the set of the first [unparseable or potentially dangerous latex formula] natural numbers so that at least two of the five numbers are consecutive. Find the remainder when [unparseable or potentially dangerous latex formula] is divided by [unparseable or potentially dangerous latex formula]

[jahu]

dont know if my logic is correct. but 2 combination problems? first you find out how many sets you can make which is 14C5, giving you 2002.

Then you find out how many dont have consecutive ones. There can be a total of 5 numbers consecutive, which is 4 attachments, and you subtract 4 from 14 which is 10. 10C5 is 252.

2002-252 = 1750

1750/1000 has a remainder of 750?

[stupidityismygam]

Yes. Here is my solution.

There are a total of [unparseable or potentially dangerous latex formula] subsets. Let's find the number of subsets that have no consecutive elements. Let [unparseable or potentially dangerous latex formula] elements of the subsets with [unparseable or potentially dangerous latex formula] Define [unparseable or potentially dangerous latex formula] The elements of [unparseable or potentially dangerous latex formula] are restricted to the set [unparseable or potentially dangerous latex formula] There is a bijection between the possible nonconsecutive [unparseable or potentially dangerous latex formula] with the number of possible [unparseable or potentially dangerous latex formula] There are [unparseable or potentially dangerous latex formula] possible [unparseable or potentially dangerous latex formula] combinations so there are the same number of nonconsecutive elements. This leaves us with [unparseable or potentially dangerous latex formula] consecutive 5-element subsets.

[jahu]

i think i need to get myself familiar with more math theory to give better solutions =P