2020 AMC 10B Problem 5

Problem 5 of 25EasierCombinatorics

How many distinguishable arrangements are there of

1

brown tile,

1

purple tile,

2

green tiles, and

3

yellow tiles in a row from left to right? (Tiles of the same color are indistinguishable.)

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Solution

There are

1+1+2+3=7

total tiles. If all seven tiles were distinct, there would be

7!

arrangements. The two green tiles are indistinguishable, and the three yellow tiles are indistinguishable, so we divide by

2!

and

3!.

Thus the number of arrangements is

\frac{7!}{2!3!}=420.

Thus, B is the correct answer.

Tagged: multiset permutations