Difference between revisions of "2022 AMC 8 Problems/Problem 8"
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==Solution 2== | ==Solution 2== | ||
− | 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 | ~hh99754539 |
Revision as of 09:58, 1 February 2022
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
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.