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 $f(5) = 1$ with the only possible sequence being $01010$ and $f(6) = 2$ with the only two possible sequences being $011010$ and $010110$ by hand, and then by the recursive formula, we have $f(19) = \boxed{\textbf{(C) }65}$ .

(Note from different author) Here are the steps if you want to see them: $f(5) = 1$ because the only possible sequence is $01010.$ $f(6) = 2$ because the two that work are $011010$ and $010110.$ $f(7) = 2$ because the two that work are $0101010$ and $0110110.$ Then, using the recursion function, we know that $f(8) = 1 + 2 = 3, f(9) = 2 + 2 = 4, f(10) = 2 + 3 = 5, f(11) = 3 + 4 = 7, f(12) = 4 + 5 = 9, f(13) = 5 + 7 = 12, f(14) = 7 + 9 = 16, f(15) = 9 + 12 = 21, f(16) = 12 + 16 = 28, f(17) = 16 + 21 = 37, f(18) = 21 + 28 = 49,$ and finally $f(19) = 28 + 37 = 65.$ So, our answer is $\boxed{65}$ ~solasky (of the note, not the solution above.)

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 ${7\choose4} = 35$ ways to arrange them.

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

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

Solution 3 (casework and blocks)

We can simplify the original problem into a problem where there are $2^{17}$ binary characters with zeros at the beginning and the end. Then, we know that we cannot have a block of 2 zeroes and a block of 3 ones. Thus, our only options are a block of $0$ s, $1$ s, and $11$ s. Now, we use casework:

Case 1: Alternating 1s and 0s. There is simply 1 way to do this: $0101010101010101010$ . Now, we note that there cannot be only one block of $11$ in the entire sequence, as there must be zeroes at both ends and if we only include 1 block, of $11$ s this cannot be satisfied. This is true for all odd numbers of $11$ blocks.

Case 2: There are 2 $11$ blocks. Using the zeroes in the sequence as dividers, we have a sample as $0110110101010101010$ . We know there are 8 places for $11$ s, which will be filled by $1$ s if the $11$ s don't fill them. This is ${8\choose2} = 28$ ways.

Case 3: Four $11$ blocks arranged. Using the same logic as Case 2, we have ${7\choose4} = 35$ ways to arrange four $11$ blocks.

Case 4: No single $1$ blocks, only $11$ blocks. There is simply one case for this, which is $0110110110110110110$ .

Adding these four cases, we have $1+28+35+1=\boxed{\textbf{(C) }65}$ as our final answer.

~Equinox8

Solution 4 (similar to #3)

Any valid sequence must start with a 0. We can then think of constructing a sequence as adding groups of terms to this 0, each ending in 0. (This is always possible, because every valid string ends in 0.) For example, we can represent the string 01011010110110 as: 0 - 10 - 110 - 10 - 110 - 110. To not have any consecutive 0s, we must have at least one 1 before the next 0. However, we cannot have 3 or more 1s before the next 0 because we cannot have 3 consecutive 1s. Consequently, we can only have one or two 1s. So we can have the groups: 10, 110.

After the initial 0, we have 18 digits left to fill in the string. Let the number of "10" blocks be x, and "110" be y. Then x and y must satisfy $2x+3y=18$ . We recognize this as a Diophantine equation. Taking $(mod 2)$ yields $y=0 (mod 2)$ . Since x and y must both be nonnegative, we get the solutions (9, 0); (6, 2); (3, 4); (0, 6). We now handle each of these cases separately.

(9, 0): Only one arrangement, namely all "01".

(6, 2): Of the 8, we choose 2 to be "001". This has 8C2=28 cases.

(3, 4): Of the 7, we choose 4 to be "001". This has 7C3=35 cases.

(0, 6): Only one arrangement, namely all "001".

Adding these, we have $1+28+35+1=65 \longrightarrow \boxed{(C)}$ . ~Math4Life2020

Video Solution

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

2019 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

Page

Toolbox

Search