Difference between revisions of "2019 AMC 10B Problems/Problem 25"
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-Solution by MagentaCobra | -Solution by MagentaCobra | ||
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| + | 2) 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 (there are 8 choose 2 ways to arrange this, or 28). | ||
| + | Option 3: four 3's and three 2's (there are 7 choose 3 ways to arrange this, or 35). | ||
| + | Option 4: six 3's (there is only 1 way to arrange this). | ||
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| + | Sum the four numbers given above: 1+28+35+1=65 \quad \boxed{C}$ | ||
==See Also== | ==See Also== | ||
Revision as of 13:42, 14 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
We can deduce that any valid sequence of length 
wil 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 remove either the last two numbers ("10") or the last three numbers ("110") and the sequence would still satisfy the given conditions.
Since 
and 
, you follow the recursion up until 
-Solution by MagentaCobra
2) 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 (there are 8 choose 2 ways to arrange this, or 28).
Option 3: four 3's and three 2's (there are 7 choose 3 ways to arrange this, or 35).
Option 4: six 3's (there is only 1 way to arrange this).
Sum the four numbers given above: 1+28+35+1=65 \quad \boxed{C}$
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. 
