2020 AMC 10B Problem 5

There are $1 + 1 + 2 + 3 = 7$ tiles, so think of $7$ empty spots in a row. Fill them in one color at a time, starting with the biggest group.

1. Yellow. Choose which $3$ of the $7$ spots are yellow. The three yellow tiles look the same, so only the choice of spots matters:
$$\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 \text{ ways}.$$
2. Green. Now $4$ spots are left. Choose which $2$ of them are green:
$$\binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6 \text{ ways}.$$
3. Brown and purple. Two spots are left. Brown goes in one and purple in the other. These tiles are different, so there are $2$ ways.
4. Each choice at one step can be combined with each choice at the next, so multiply:
$$35 \times 6 \times 2 = 420.$$

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

When some objects look the same, you only have to decide where each color goes. Choosing the spots for one color at a time handles this, because choosing spots does not care about order inside a group. The count does not depend on which color you place first.

Pretend for a moment that all $7$ tiles are different. Then there are $7! = 5040$ orders. Each real arrangement has been counted many times: the two green tiles can swap ($2! = 2$ ways) and the three yellow tiles can be shuffled ($3! = 6$ ways) without changing what you see. So each arrangement was counted $2 \times 6 = 12$ times, and the true count is $5040 \div 12 = 420$.

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 $\boxed{\textbf{(B)}\ 420}$.