If n items are arranged two at a time, the number obtained is 20. Find the value of n.
\(^{n}P_{2} = \frac{n!}{(n - 2)!} = 20 \)
\(\frac{n(n - 1)(n - 2)!}{(n - 2)!} = 20\)
\(n(n - 1) = 20 \implies n^{2} - n - 20 = 0\)
\(n^{2} - 5n + 4n - 20 = 0\)
\(n(n - 5) + 4(n - 5) = 0\)
\(n = \text{5 or -4}\)
\(n = 5\)
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