2015 AMC 10A Problem 10

Below is the professionally curated solution for Problem 10 of the 2015 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2015 AMC 10A solutions, or check the answer key.

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Concepts:arrangements with restrictionscasework

Difficulty rating: 1370

10.

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

00

11

22

33

44

Solution:

The initial step would be to choose a letter that could work and build from there.

For instance, if we begin with a,a, the only letters that can be placed next to it are cc or d.d.

From here, we cannot proceed further since the combinations acbdacbd and acdbacdb are not allowed due to the presence of cbcb and cd,cd, respectively.

The same issue arises if we start with d,d, as a bb would have to be placed in the middle and end up being adjacent to either an aa or a c.c.

On the other hand, if we start with a b,b, the next letter would have to be d,d, and after that, an aa and a cc can be placed, making this arrangement possible.

The same methodology holds true for starting with a c.c. This gives us a total of 22 rearrangements.

Thus, C is the correct answer.

Problem 10 in Other Years