Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Arranging in column.

## Arranging in column – AIME I, 1990

In a shooting match, eight clay targets are arranged in two hanging columns of three targets each and one column of two targets. A marks man 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?

- is 107
- is 560
- is 840
- cannot be determined from the given information

**Key Concepts**

Integers

Arrangement

Algebra

## Check the Answer

But try the problem first…

Answer: is 560.

AIME I, 1990, Question 8

Combinatorics by Brualdi

## Try with Hints

First hint

Let the columns be labelled A,B and C such that first three targets are A, A and A the next three being B, B and B and the next being C and C in which we consider the string AAABBBCC.

Second Hint

Since the arrangement of the strings is one-one correspondence and onto to the order of shooting for example first A is shot first, second A is shot second, third A is shot third, first B is shot fourth, second B is shot fifth, third B is shot sixth, first C is shot seventh, second C is shot eighth,

or, here arrangement of the strings is bijective to the order of the shots taken

Final Step

the required answer is the number of ways to arrange the letters which is \(\frac{8!}{3!3!2!}\)=560.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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