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. | ||
− | If the rules are followed, in how many different | + | If the rules are followed, in how many different orders can the eight targets be broken? |
+ | |||
+ | [[File:1990 AIME Problem 8.png|center|160px]] | ||
== Solution == | == 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 <math>L,M,</math> and <math>R,</math> respectively. We consider the string <math>LLLMMRRR:</math> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ~Azjps (Solution) | ||
+ | |||
+ | ~MRENTHUSIASM (Revision) | ||
+ | |||
+ | == Remark == | ||
+ | We can count the letter arrangements of <math>LLLMMRRR</math> in two ways: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>There are <math>8!</math> ways to arrange <math>8</math> distinguishable letters. However: <p> | ||
+ | <ul style="list-style-type:square;"> | ||
+ | <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> | ||
+ | <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> | ||
+ | <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> | ||
+ | </ul> | ||
+ | By the Multiplication Principle, we have counted each distinct arrangement of <math>LLLMMRRR</math> for <math>3!\cdot2!\cdot3!</math> times. <p> | ||
+ | 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> | ||
+ | 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> | ||
+ | <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> | ||
+ | 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> | ||
+ | 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> | ||
+ | By the Multiplication Principle, the answer is <math>\binom83\binom52\binom33=560.</math> | ||
+ | </li><p> | ||
+ | </ol> | ||
+ | ~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 == | == See also == |
Latest revision as of 04:24, 4 November 2022
Contents
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
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 and
respectively. We consider the string
Since the letter arrangements of and the shooting orders have one-to-one correspondence, we count the letter arrangements:
~Azjps (Solution)
~MRENTHUSIASM (Revision)
Remark
We can count the letter arrangements of in two ways:
- There are
ways to arrange
distinguishable letters. However:
- Since there are
indistinguishable
's, we have counted each distinct arrangement of the
's for
times.
- Since there are
indistinguishable
's, we have counted each distinct arrangement of the
's for
times.
- Since there are
indistinguishable
's, we have counted each distinct arrangement of the
's for
times.
for
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:
- Since there are
- First, we have
ways to choose any
from the
positions for the
's.
Next, we have
ways to choose any
from the
remaining positions for the
's.
Finally, we have
way to choose
from the
remaining positions for the
's.
By the Multiplication Principle, the answer is
~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) | ||
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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