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2020 AMC 10B Problem 5

2020 AMC 10B Problem 5AMC 10 Step by Step

Name: Date: September 13, 2026

Basic Counting · difficulty 2 of 5 · about 2 min · position P1-10

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.)

A)
$210$
B)
$420$
C)
$630$
D)
$840$
E)
$1050$

Answer

Key insight

Arrange 7 tiles as if distinct (7!), then divide by 2! and 3! for the identical green and yellow tiles.

Solution

There are $1 + 1 + 2 + 3 = 7$ tiles. If all were different there would be $7! = 5040$ orderings. Swapping the two green tiles among themselves ( $2!$ ways) or permuting the three yellow tiles ( $3!$ ways) does not change the visible arrangement, so each distinguishable arrangement was counted $2! \cdot 3! = 12$ times:
$\frac{7!}{2!\,3!} = \frac{5040}{12} = 420.$

The answer is $\boxed{\textbf{(B)}\ 420}$ .

Why this works

Permutations of a multiset: $\dfrac{n!}{n_1!\,n_2!\cdots}$ where the $n_i$ are the group sizes. The division corrects the overcount from treating identical objects as labeled. This formula is exactly the multinomial coefficient and shows up constantly in word-rearrangement and tile problems.

Alternative approach

Place colors one group at a time: choose $3$ of the $7$ positions for yellow ( $\binom73 = 35$ ), then $2$ of the remaining $4$ for green ( $\binom42 = 6$ ), then the last two positions get brown and purple in $2$ ways: $35 \cdot 6 \cdot 2 = 420$ .

The trap

Forgetting to divide by 3! for the yellow tiles, or dividing by 2 and 3 instead of 2! and 3!.

Common mistakes

Forgetting to divide by 3! for the yellow tiles, or dividing by 2 and 3 instead of 2! and 3!.
Using $\binom{7}{2}\binom{7}{3}$ without removing already-filled positions, or treating the four colors as four objects ( $4! = 24$ ).

Techniques

Set up the equation/formula and compute; no special trick needed