During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

# Difference between revisions of "2020 AMC 10B Problems/Problem 5"

## Problem

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.)

$\textbf{(A)}\ 210 \qquad\textbf{(B)}\ 420 \qquad\textbf{(C)}\ 630 \qquad\textbf{(D)}\ 840 \qquad\textbf{(E)}\ 1050$

## Solution

Let's first find how many possibilities there would be if they were all distinguishable, then divide out the ones we overcounted.

There are $7!$ ways to order $7$ objects. However, since there's $3!=6$ ways to switch the yellow tiles around without changing anything (since they're indistinguishable) and $2!=2$ ways to order the green tiles, we have to divide out these possibilities.

$\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}$ ~quacker88

## Solution

We can repeat chooses extensively to find the answer. There are $7$ choose $3$ ways to arrange the brown tiles which is $35$. Then from the remaining tiles there are $4$ choose $2=6$ ways to arrange the red tiles. And now from the remaining two tiles and two slots we can see there are two ways to arrange the purple and brown tiles, giving us an answer of $35*6*2=420$ $\frac{7!}{6\cdot2}=\boxed{\textbf{(B) }420}$ - noahdavid

~IceMatrix