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 and respectively.

Consider the string 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

~Azjps (Solution)

~MRENTHUSIASM (Revision)

Remark

We can count the letter arrangements of in two ways:

1. There are ways to arrange distinguishable letters. However:
   • Since there are indistinguishable 's, we have counted each distinct arrangement times.
   • Since there are indistinguishable 's, we have counted each distinct arrangement times.
   • Since there are indistinguishable 's, we have counted each distinct arrangement times.

   By the Multiplication Principle, we have counted each distinct arrangement times.

   As shown in the Solution section, we use division to fix the overcount. The answer is

   Alternatively, we can use a multinomial coefficient to obtain the answer directly:

2. First, we have ways to choose any from the spots for the 's.

   Next, we have ways to choose any from the remaining spots for the 's.

   Finally, we have way to choose from the remaining spots for the 's.

   By the Multiplication Principle, the answer is

~MRENTHUSIASM

Video Solution

https://www.youtube.com/watch?v=NGfMLCRUs3c&t=7s ~ MathEx

See also

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.