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Difference between revisions of "2015 AMC 10A Problems/Problem 10"

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<cmath>1 + 1 = \boxed{\textbf{(C)}\ 2}.</cmath>
 
<cmath>1 + 1 = \boxed{\textbf{(C)}\ 2}.</cmath>
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== Video Solution ==
 +
https://youtu.be/3MiGotKnC_U?t=1509
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~ ThePuzzlr
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 11:25, 19 October 2021

Problem

How many rearrangements of $abcd$ are there in which no two adjacent letters are also adjacent letters in the alphabet? For example, no such rearrangements could include either $ab$ or $ba$.

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$

Solution

The first thing one would want to do is place a possible letter that works and then stem off of it. For example, if we start with an $a$, we can only place a $c$ or $d$ next to it. Unfortunately, after that step, we can't do too much, since:

$acbd$ is not allowed because of the $cb$, and $acdb$ is not allowed because of the $cd$.

We get the same problem if we start with a $d$, since a $b$ will have to end up in the middle, causing it to be adjacent to an $a$ or $c$.

If we start with a $b$, the next letter would have to be a $d$, and since we can put an $a$ next to it and then a $c$ after that, this configuration works. The same approach applies if we start with a $c$.

So the solution must be the two solutions that were allowed, one starting from a $b$ and the other with a $c$, giving us:

\[1 + 1 = \boxed{\textbf{(C)}\ 2}.\]

Video Solution

https://youtu.be/3MiGotKnC_U?t=1509

~ ThePuzzlr

Video Solution

https://youtu.be/0W3VmFp55cM?t=1055

~ pi_is_3.14

Video Solution

https://youtu.be/8sTQIX4YJ6s

~savannahsolver

See Also

2015 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
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

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