Discussion
If \({}^{n}P_{5} = 42\,{}^{n}P_{3}\) then \(n\) is equal to
Answer: E
\(n = 10\).
Use \({}^{n}P_{r} = n(n-1)\cdots(n-r+1)\). The first three factors are common to both sides and cancel:
\[ \begin{aligned} n(n-1)(n-2)(n-3)(n-4) &= 42\,n(n-1)(n-2) \\ (n-3)(n-4) &= 42 = 7 \times 6 \\ n - 3 &= 7, \quad n = 10 \end{aligned} \]
Check: \({}^{10}P_{5} = 30240\) and \({}^{10}P_{3} = 720\), and \(720 \times 42 = 30240\). The smaller values 3, 5 and 7 make \({}^{n}P_{5}\) zero or too small, and 12 gives \((9)(8) = 72\), not 42.
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