2017 AMC 10A Problem 19

Below is the professionally curated solution for Problem 19 of the 2017 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 10A solutions, or check the answer key.

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Concepts:arrangements with restrictionscasework

Difficulty rating: 1660

19.

Alice refuses to sit next to either Bob or Carla. Derek refuses to sit next to Eric. How many ways are there for the five of them to sit in a row of 55 chairs under these conditions?

1212

1616

2828

3232

4040

Solution:

If Alice sits on an end, then the person next to her cannot be Bob or Carla. This means that it must be Derek or Eric.

WLOG, let the person be Eric. Then the person next to Eric has to be Bob or Carla. After that there are no more restrictions.

This gives us a total of 2222=16. 2 \cdot 2 \cdot 2 \cdot 2 = 16.

The first 22 is for both edges. The second 22 is for Derek or Eric. The third 22 is for Bob or Carla. The final 22 is just for the 22 people that are remaining.

Otherwise, let Alice be in one of the three non-end seats. Then the two people next to her have to be Derek and Eric. Bob and Carla are forced to be in the last 22 seats.

There are 33 choices for Alice's seat. The side on which Derek sits has 22 options, and then there are 22 options for where Bob and Carla go.

This gives us 322=12 3 \cdot 2 \cdot 2 = 12 configurations.

Therefore, there are a total of 12+16=2812 + 16 = 28 total seating arrangements.

Thus, C is the correct answer.

Problem 19 in Other Years