# Difference between revisions of "1990 AIME Problems/Problem 8"

## Problem

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marksman is to break all the targets according to the following rules:

1) The marksman first chooses a column from which a target is to be broken.

2) The marksman must then break the lowest remaining target in the chosen column.

If the rules are followed, in how many different orders can the eight targets be broken?

## Solution

From left to right, suppose that the columns are labeled $L,M,$ and $R,$ respectively.

Consider the string $LLLMMRRR.$ Since the set of all letter arrangements is bijective to the set of all shooting orders, the answer is the number of letter arrangements, which is $\frac{8!}{3!\cdot2!\cdot3!} = \boxed{560}.$

~Azjps (Solution)

~MRENTHUSIASM (Revision)

## Remark

We can count the letter arrangements of $LLLMMRRR$ in two ways:

1. There are $8!$ ways to arrange $8$ distinguishable letters. However:

• Since there are $3$ indistinguishable $L$'s, we have counted each distinct arrangement $3!$ times.
• Since there are $2$ indistinguishable $M$'s, we have counted each distinct arrangement $2!$ times.
• Since there are $3$ indistinguishable $R$'s, we have counted each distinct arrangement $3!$ times.
By the Multiplication Principle, we have counted each distinct arrangement $3!\cdot2!\cdot3!$ times.

As shown in the Solution section, we use division to fix the overcount. The answer is $\frac{8!}{3!\cdot2!\cdot3!} = 560.$

Alternatively, we can use a multinomial coefficient to obtain the answer directly: $\binom{8}{3,2,3}=\frac{8!}{3!\cdot2!\cdot3!} = 560.$

2. First, we have $\binom83$ ways to choose any $3$ from the $8$ spots for the $L$'s.

Next, we have $\binom52$ ways to choose any $2$ from the $5$ remaining spots for the $M$'s.

Finally, we have $\binom33$ way to choose $3$ from the $3$ remaining spots for the $R$'s.

By the Multiplication Principle, the answer is $\binom83\binom52\binom33=560.$

~MRENTHUSIASM

## Video Solution

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 