# Difference between revisions of "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, from the given restrictions, that any valid sequence of length $n$ will start with a $0$ followed by either $10$ or $110$. Thus we can define a recursive function $f(n) = f(n-3) + f(n-2)$, where $f(n)$ is the number of valid sequences of length $n$.

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

It is easy to find $f(5) = 1$ and $f(6) = 2$ by hand, and then by the recursive formula, we have $f(19) = \boxed{\textbf{(C) }65}$.

## Solution 2 (casework)

After any particular $0$, the next $0$ in the sequence must appear exactly $2$ or $3$ positions down the line. In this case, we start at position $1$ and end at position $19$, i.e. we move a total of $18$ positions down the line. Therefore, we must add a series of $2$s and $3$s to get $18$. There are a number of ways to do this:

Case 1: nine $2$s - there is only $1$ way to arrange them.

Case 2: two $3$s and six $2$s - there are ${8\choose2} = 28$ ways to arrange them.

Case 3: four $3$s and three $2$s - there are ${7\choose3} = 35$ ways to arrange them.

Case 4: six $3$s - there is only $1$ way to arrange them.

Summing the four cases gives $1+28+35+1=\boxed{\textbf{(C) }65}$.

## Video Solution

For those who want a video solution: https://youtu.be/VamT49PjmdI