2021 JMPSC Accuracy Problems/Problem 5

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Problem

Let $n!=n \cdot (n-1) \cdot (n-2) \cdots 2 \cdot 1$ for all positive integers $n$. Find the value of $x$ that satisfies \[\frac{5!x}{2022!}=\frac{20}{2021!}.\]

Solution

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