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 s and s of length are there that begin with a , end with a , contain no two consecutive s, and contain no three consecutive s?

Solution 1 (Recursion)

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

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

and , so you follow the recursion up until

~Solution by MagentaCobra

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 ( ways to arrange this).

Option 3: four 3's and three 2's ( 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

~Solution by mxnxn

Solution3

That problem have 3 different cases like 0 or 0->1 or 1->1 If one sequence last two is 1->1 than next well be 0 If one sequense last two is 0->1 then next can be 0 or 1 If one sequense last one is 0 then next well be 1

       1       2       3       4       5       6       7       8       9 

0 1 0 1 1 1 2 2 3 4 0 1 0 1 0 1 1 1 2 2 3 1 1 0 0 1 0 1 1 1 2 2 10 11 12 13 14 15 16 17 18 0 5 7 9 12 16 21 28 37 49 0 1 4 5 7 9 12 16 21 28 37 1 1 3 4 5 7 9 12 16 21 28 But last one is 0 so just 1->1->0 or 0->1->1 well be right->answer is 37+28=65(C)

~Solution by rayfunmath

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.