2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 25 Hard

How many sequences of 0\rm s and 1\rm s of length 19 are there that begin with a 0, end with a 0, contain no two consecutive 0\rm s, and contain no three consecutive 1\rm s? (2019 AMC 10B Problem, Question#25)

  • A.

    55

  • B.

    60

  • C.

    65

  • D.

    70

  • E.

    75

Answer:C

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)=(\text{C})65.

After any particular 0 , the next  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\rm s and \rm 3s to get 18. There are a number of ways to do this:

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

Case 2: two 3\rm s and six \rm 2s - there are \left( \begin{array}{l} {8}\\{2} \end{array}\right)=28 ways to arrange them.

Case 3: four 3\rm s and three 2\rm s - there are \left( \begin{array}{l} {7}\\{3} \end{array}\right)=35 ways to arrange them.

Case 4: six 3\rm s - there is only 1 way to arrange them. Summing the four cases gives 1+28+35+1=(\text{C})65.