# 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 $0$s and $1$s of length $19$ are there that begin with a $0$, end with a $0$, contain no two consecutive $0$s, and contain no three consecutive $1$s?

$\textbf{(A) }55\qquad\textbf{(B) }60\qquad\textbf{(C) }65\qquad\textbf{(D) }70\qquad\textbf{(E) }75$

## Solution 1 (Recursion)

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

$f(n) = f(n-3) + f(n-2)$

This is because for any valid sequence of length $n$, you can append either "10" or "110" and the resulting sequence would still satisfy the given conditions.

$f(5) = 1$ and $f(6) = 2$, so you follow the recursion up until $f(19) = 65 \quad \boxed{C}$

## Solution 2 (Casework)

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 (${8\choose2} =28$ ways to arrange this).

Option 3: four 3's and three 2's (${7\choose3}=35$ ways to arrange this).

Option 4: six 3's (there is only 1 way to arrange this).

Sum the four numbers given above: 1+28+35+1=65

## Solution3

first have 1 choose second have 1 choose 2 2 3 4 5 7 9 12 16 21 28 37 49 86 65 last is 0 at the last we see the answer is 65(C) (Can someone please edit this?)