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 $110$ . Let $f(n)$ be the number of valid (meaning: the sequence contains 0s and 1s, starts and ends with both 0, and there are no two consecutive 0s and no three consecutive 1s) sequences of length $n$ .

Then we can define a recursive function $f(n) = f(n-3) + f(n-2)$ , with $n \ge 3$ (because otherwise, the sequence would contain only 0s and this is not allowed due to the give conditions).

We derived the recursive function, since for any valid sequence of length $n$ , you can append either $10$ or $110$ to return to the starting position, 0, and the resulting sequence will still satisfy the given conditions.

It is easy to find $f(3) = 1$ since the only possible valid sequence is $010$ . $f(4)=1$ since the only possible valid sequence is $0110$ . $f(5)=1$ since the only possible valid sequence is $01010$ .

The recursive sequence is then as follows:

$f(3)=1$ $f(4)=1$ $f(5) = 1$ $f(6) = 1 + 1 = 2$ $f(7) = 1 + 1 = 2$ $f(8) = 1 + 2 = 3$ $f(9) = 2 + 2 = 4$ $f(10) = 2 + 3 = 5$ $f(11) = 3 + 4 = 7$ $f(12) = 4 + 5 = 9$ $f(13) = 5 + 7 = 12$ $f(14) = 7 + 9 = 16$ $f(15) = 9 + 12 = 21$ $f(16) = 12 + 16 = 28$ $f(17) = 16 + 21 = 37$ $f(18) = 21 + 28 = 49$ $f(19) = 28 + 37 = 65$

So, our answer is $\boxed{\text{\bf{(C)} } 65}$ .

Contributors:

~Original Author

~solasky

~BakedPotato66

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 three or more $1$ s before the next $0$ because we cannot have three consecutive $1$ s. Consequently, we can only have one or two $1$ s.

So we can have the groups: $10$ and $110$ .

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 $\pmod{2}$ yields $y=0 \pmod{2}$ . Since $x$ and $y$ must both be nonnegative, we get the solutions $(9, 0)$ , $(6, 2)$ , $(3, 4)$ , and $(0, 6)$ . We now handle each of these cases separately.

$(9, 0)$ : Only one arrangement, namely all $10$ s.

$(6, 2)$ : We have 6 groups of $11$ , and $2$ groups of $110$ . This has $\binom{6+2}{2}=28$ cases.

$(3, 4)$ : This means we have 3 groups of $10$ , and 4 groups of $110$ . This has $\binom{3+4}{3}=35$ cases.

$(0, 6)$ : Only one arrangement, namely all $110$ .

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

~edited by alpha_2 for spelling and and typos ( $001$ instead of $110$ )

Video Solution

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

2019 AMC 10B (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2019 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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