Difference between revisions of "2019 AMC 10B Problems/Problem 25"
Rayfunmath (talk | contribs) |
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first have 1 choose | first have 1 choose | ||
second have 1 choose | second have 1 choose | ||
− | + | 2 2 3 4 5 7 9 12 16 21 28 37 49 86 65 last is 0 | |
at the last we see the answer is 65(C) | at the last we see the answer is 65(C) | ||
~Solution by rayfunmath | ~Solution by rayfunmath |
Revision as of 04:33, 15 February 2019
- 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
first have 1 choose second have 1 choose 2 2 3 4 5 7 9 12 16 21 28 37 49 86 65 last is 0 at the last we see the answer is 65(C) ~Solution by rayfunmath
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.