Discussion

If \({}^{n}P_{5} = 360360\), then \({}^{n}C_{5}\) =

Answer: E

\({}^{n}C_{5}\) = 3003.

Permutations and combinations differ by the 5! orderings of the chosen items:

\[ \begin{aligned} {}^{n}C_{5} &= \frac{{}^{n}P_{5}}{5!} \\ &= \frac{360360}{120} = 3003 \end{aligned} \]

Here n = 15, since \(15 \times 14 \times 13 \times 12 \times 11 = 360360\), and \({}^{15}C_{5} = 3003\).

The step works because each combination of 5 objects can be arranged in 5! = 120 orders, and each order is a different permutation.

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