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

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Of the 8 letters, 3 of them are <math>A</math>s, making <math>{8\choose3}</math> possibilities. Of the remaining 5, 3 are <math>B</math>s, making <math>{5\choose3}</math> possibilities. The positions of the 2 <math>C</math>s are then fixed. Thus, there are <math>{8\choose3} \cdot {5\choose3} = 560</math> ways of shooting all of the targets.
 
Of the 8 letters, 3 of them are <math>A</math>s, making <math>{8\choose3}</math> possibilities. Of the remaining 5, 3 are <math>B</math>s, making <math>{5\choose3}</math> possibilities. The positions of the 2 <math>C</math>s are then fixed. Thus, there are <math>{8\choose3} \cdot {5\choose3} = 560</math> ways of shooting all of the targets.
  
Alternatively, the number of ways to arrange 3 <math>A</math>s, 3 <math>B</math>s, and 2 <math>C</math>s in a string of 8 letters is equal to <math>\frac{8!}{3! \cdot 3! \cdot 2!} = 560</math>.
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Alternatively, the number of ways to arrange 3 <math>A</math>s, 3 <math>B</math>s, and 2 <math>C</math>s in a string of 8 letters is equal to <math>\frac{8!}{3! \cdot 3! \cdot 2!} = \boxed{560}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 01:00, 8 September 2011

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

Suppose that the columns are labeled $A$, $B$, and $C$. The question is asking for the number of ways to shoot at the bottom target of the columns, or the number of ways to arrange 3 $A$s, 3 $B$s, and 2 $C$s in a string of 8 letters.

Of the 8 letters, 3 of them are $A$s, making ${8\choose3}$ possibilities. Of the remaining 5, 3 are $B$s, making ${5\choose3}$ possibilities. The positions of the 2 $C$s are then fixed. Thus, there are ${8\choose3} \cdot {5\choose3} = 560$ ways of shooting all of the targets.

Alternatively, the number of ways to arrange 3 $A$s, 3 $B$s, and 2 $C$s in a string of 8 letters is equal to $\frac{8!}{3! \cdot 3! \cdot 2!} = \boxed{560}$.

See also

1990 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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