2020 AMC 8 Problem 10

Below is the video solution and professionally curated solution for Problem 10 of the 2020 AMC 8, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 8 solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:arrangements with restrictionscomplementary counting

Difficulty rating: 960

10.

Zara has a collection of 44 marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this?

66

88

1212

1818

2424

Video solution:
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Written solution:

There are 4!=244!=24 total arrangements of the marbles.

If the Steelie and Tiger are adjacent, treat them as one block. Then there are 3!3! ways to arrange the block with the other two marbles, and 22 orders inside the block, for 3!2=123!\cdot2=12 adjacent arrangements.

Therefore, 2412=1224-12=12 arrangements keep the Steelie and Tiger separated.

Thus, the correct answer is C.

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