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

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2) The marksman must then break the lowest remaining target in the chosen column.  
 
2) The marksman must then break the lowest remaining target in the chosen column.  
  
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If the rules are followed, in how many different [[order]]s can the eight targets be broken?  
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If the rules are followed, in how many different orders can the eight targets be broken?
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[[File:1990 AIME Problem 8.png|center|160px]]
  
 
== Solution ==
 
== Solution ==
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Suppose that the columns are labeled <math>A</math>, <math>B</math>, and <math>C</math>. 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 <math>A</math>s, 3 <math>B</math>s, and 2 <math>C</math>s in a string of 8 letters.
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Clearly, the marksman must shoot the left column three times, the middle column two times, and the right column three times.
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 +
From left to right, suppose that the columns are labeled <math>L,M,</math> and <math>R,</math> respectively. We consider the string <math>LLLMMRRR:</math>
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Since the letter arrangements of <math>LLLMMRRR</math> and the shooting orders have one-to-one correspondence, we count the letter arrangements: <cmath>\frac{8!}{3!\cdot2!\cdot3!} = \boxed{560}.</cmath>
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~Azjps (Solution)
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 +
~MRENTHUSIASM (Revision)
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 +
== Remark ==
 +
We can count the letter arrangements of <math>LLLMMRRR</math> in two ways:
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<ol style="margin-left: 1.5em;">
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  <li>There are <math>8!</math> ways to arrange <math>8</math> distinguishable letters. However: <p>
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<ul style="list-style-type:square;">
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<li>Since there are <math>3</math> indistinguishable <math>L</math>'s, we have counted each distinct arrangement of the <math>L</math>'s for <math>3!</math> times. </li><p>
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<li>Since there are <math>2</math> indistinguishable <math>M</math>'s, we have counted each distinct arrangement of the <math>M</math>'s for <math>2!</math> times. </li><p>
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<li>Since there are <math>3</math> indistinguishable <math>R</math>'s, we have counted each distinct arrangement of the <math>R</math>'s for <math>3!</math> times. </li><p>
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</ul>
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By the Multiplication Principle, we have counted each distinct arrangement of <math>LLLMMRRR</math> for <math>3!\cdot2!\cdot3!</math> times. <p>
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As shown in the <b>Solution</b> section, we use division to fix the overcount. The answer is <math>\frac{8!}{3!\cdot2!\cdot3!} = 560.</math> <p>
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Alternatively, we can use a multinomial coefficient to obtain the answer directly: <math>\binom{8}{3,2,3}=\frac{8!}{3!\cdot2!\cdot3!} = 560.</math> </li><p>
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  <li>First, we have <math>\binom83</math> ways to choose any <math>3</math> from the <math>8</math> positions for the <math>L</math>'s.<p>
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Next, we have <math>\binom52</math> ways to choose any <math>2</math> from the <math>5</math> remaining positions for the <math>M</math>'s.<p>
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Finally, we have <math>\binom33</math> way to choose <math>3</math> from the <math>3</math> remaining positions for the <math>R</math>'s.<p>
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By the Multiplication Principle, the answer is <math>\binom83\binom52\binom33=560.</math>
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</li><p>
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</ol>
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~MRENTHUSIASM
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== Video Solution by OmegaLearn ==
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https://youtu.be/RldWnL4-BfI?t=448
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 +
~ pi_is_3.14
  
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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.
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==Video Solution==
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https://www.youtube.com/watch?v=NGfMLCRUs3c&t=7s ~ MathEx
  
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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!} = 560</math>.
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==Video Solution==
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https://www.youtube.com/watch?v=eq50QuGasg0
  
 
== See also ==
 
== See also ==
Line 19: Line 56:
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 03:24, 4 November 2022

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?

1990 AIME Problem 8.png

Solution

Clearly, the marksman must shoot the left column three times, the middle column two times, and the right column three times.

From left to right, suppose that the columns are labeled $L,M,$ and $R,$ respectively. We consider the string $LLLMMRRR:$

Since the letter arrangements of $LLLMMRRR$ and the shooting orders have one-to-one correspondence, we count the letter arrangements: \[\frac{8!}{3!\cdot2!\cdot3!} = \boxed{560}.\]

~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 of the $L$'s for $3!$ times.

    • Since there are $2$ indistinguishable $M$'s, we have counted each distinct arrangement of the $M$'s for $2!$ times.

    • Since there are $3$ indistinguishable $R$'s, we have counted each distinct arrangement of the $R$'s for $3!$ times.

    By the Multiplication Principle, we have counted each distinct arrangement of $LLLMMRRR$ for $3!\cdot2!\cdot3!$ 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 $8$ positions for the $L$'s.

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

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

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

~MRENTHUSIASM

Video Solution by OmegaLearn

https://youtu.be/RldWnL4-BfI?t=448

~ pi_is_3.14

Video Solution

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

Video Solution

https://www.youtube.com/watch?v=eq50QuGasg0

See also

1990 AIME (Problems • Answer Key • Resources)
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Problem 7 		Followed by
Problem 9
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