Discussion
If \(n_{P_{5}} = 6720\) and \((n - 1)_{P_{4}} = 840\) , then \(n_{P_{3}}\) is equal to
Answer: E
\({}^{n}P_{3}\) is 336.
The paper's \(n_{P_{5}}\) means \({}^{n}P_{5}\). Since \({}^{n}P_{r} = n \cdot {}^{n-1}P_{r-1}\):
\[ \begin{aligned} n &= \frac{{}^{n}P_{5}}{{}^{n-1}P_{4}} = \frac{6720}{840} = 8 \\ {}^{8}P_{3} &= 8 \times 7 \times 6 = 336 \end{aligned} \]
Check: \({}^{8}P_{5} = 8 \times 7 \times 6 \times 5 \times 4 = 6720\). The other options are not products of three consecutive integers starting from 8.
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