Difference between revisions of "2019 AMC 10B Problems/Problem 25"

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If one sequense last two is 0->1 then next can be 0 or 1
 
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
 
If one sequense last one is 0 then next well be 1
1 2 3 4 5 6 7 8 9
+
        1       2       3       4       5       6       7       8       9  
0 1 0 1 1 1 2 2 3 4
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0
0->1 0 1 0 1 1 1 2 2 3
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0->1
1->1 0 0 1 0 1 1 1 2 2
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1->1  
 
 
 
10 11 12 13 14 15 16 17 18
 
10 11 12 13 14 15 16 17 18
 
0 5 7 9 12 16 21 28 37 49
 
0 5 7 9 12 16 21 28 37 49

Revision as of 05:20, 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 $0$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 that any valid sequence of length $n$ will start with a 0 followed by either "10" or "110". Because of this, we can define a recursive function:

$f(n) = f(n-3) + f(n-2)$

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

$f(5) = 1$ and $f(6) = 2$, so you follow the recursion up until $f(19) = 65 \quad \boxed{C}$

~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 (${8\choose2} =28$ ways to arrange this).

Option 3: four 3's and three 2's (${7\choose3}=35$ 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 0->1 1->1 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

2019 AMC 10B (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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