2019 AMC 10B Problems/Problem 25

The following problem is from both the 2019 AMC 10B #25 and 2019 AMC 12B #23, so both problems redirect to this page.

Problem

How many sequences of s and s of length are there that begin with a , end with a , contain no two consecutive s, and contain no three consecutive s?

Solution

We can deduce that any valid sequence of length wil start with a 0 followed by either "10" or "110". Because of this, we can define a recursive function:

This is because for any valid sequence of length , you can remove either the last two numbers ("10") or the last three numbers ("110") and the sequence would still satisfy the given conditions.

Since and , you follow the recursion up until

-Solution by MagentaCobra

2) After any given zero, the next zero must appear exactly two or three spots down the line. And we started at position 1 and ended at position 19, so we moved over 18. Therefore, we must add a series of 2's and 3's to get 18. How can we do this? Option 1: nine 2's (there is only 1 way to arrange this). Option 2: two 3's and six 2's (there are 8 choose 2 ways to arrange this, or 28). Option 3: four 3's and three 2's (there are 7 choose 3 ways to arrange this, or 35). Option 4: six 3's (there is only 1 way to arrange this).

Sum the four numbers given above: 1+28+35+1=65 \quad \boxed{C}$

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.