Difference between revisions of "2022 AMC 8 Problems/Problem 8"
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The original expression becomes <cmath>\frac{20!}{22!/2!} = \frac{20! \cdot 2!}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \boxed{\textbf{(B) } \frac{1}{231}}.</cmath> | The original expression becomes <cmath>\frac{20!}{22!/2!} = \frac{20! \cdot 2!}{22!} = \frac{20! \cdot 2}{20! \cdot 21 \cdot 22} = \frac{2}{21 \cdot 22} = \frac{1}{21 \cdot 11} = \boxed{\textbf{(B) } \frac{1}{231}}.</cmath> | ||
+ | ~hh99754539 | ||
+ | |||
+ | ==Video Solution by Math-X (First understand the problem!!!)== | ||
+ | https://youtu.be/oUEa7AjMF2A?si=FHon9G492KY6Ecf3&t=996 | ||
+ | |||
+ | ~Math-X | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING!!!)== | ||
+ | https://youtu.be/vr7hf7IK2po | ||
+ | |||
+ | ~Education, the Study of Everything | ||
− | |||
==Video Solution== | ==Video Solution== | ||
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=565 | https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=565 | ||
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~savannahsolver | ~savannahsolver | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ITVgbv6MoVA | ||
+ | |||
+ | ~harungurcan | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2022|num-b=7|num-a=9}} | {{AMC8 box|year=2022|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 13:40, 23 November 2023
Contents
Problem
What is the value of
Solution 1
Note that common factors (from to inclusive) of the numerator and the denominator cancel. Therefore, the original expression becomes
~MRENTHUSIASM
Solution 2
The original expression becomes ~hh99754539
Video Solution by Math-X (First understand the problem!!!)
https://youtu.be/oUEa7AjMF2A?si=FHon9G492KY6Ecf3&t=996
~Math-X
Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
Video Solution
https://www.youtube.com/watch?v=Ij9pAy6tQSg&t=565
~Interstigation
Video Solution
https://youtu.be/1xspUFoKDnU?t=176
~STEMbreezy
Video Solution
~savannahsolver
Video Solution
~harungurcan
See Also
2022 AMC 8 (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.