Difference between revisions of "1990 AIME Problems/Problem 8"
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<li>There are <math>8!</math> ways to arrange <math>8</math> distinguishable letters. However: <p> | <li>There are <math>8!</math> ways to arrange <math>8</math> distinguishable letters. However: <p> | ||
<ul style="list-style-type:square;"> | <ul style="list-style-type:square;"> | ||
− | <li>Since there are <math>3</math> indistinguishable <math>L</math>'s, we have counted each distinct arrangement <math>3!</math> times. </li><p> | + | <li>Since there are <math>3</math> indistinguishable <math>L</math>'s, we have counted each distinct arrangement of the <math>L</math>'s <math>3!</math> times. </li><p> |
− | <li>Since there are <math>2</math> indistinguishable <math>M</math>'s, we have counted each distinct arrangement <math>2!</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 <math>2!</math> times. </li><p> |
− | <li>Since there are <math>3</math> indistinguishable <math>R</math>'s, we have counted each distinct arrangement <math>3!</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 <math>3!</math> times. </li><p> |
</ul> | </ul> | ||
− | By the Multiplication Principle, we have counted each distinct arrangement <math>3!\cdot2!\cdot3!</math> times. <p> | + | By the Multiplication Principle, we have counted each distinct arrangement of <math>LLLMMRRR</math> <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> | 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> | 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> |
Revision as of 14:32, 19 June 2021
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:
- There are ways to arrange distinguishable letters. However:
- Since there are indistinguishable 's, we have counted each distinct arrangement of the 's times.
- Since there are indistinguishable 's, we have counted each distinct arrangement of the 's times.
- Since there are indistinguishable 's, we have counted each distinct arrangement of the 's 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:
- 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
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.