Discussion

If \({}^{11}P_{r} = 7920\) , then the value of \(r\) is equal to

Answer: D

\(r = \) 4.

\({}^{11}P_{r}\) is the product of \(r\) consecutive integers counting down from 11:

\[ \begin{aligned} {}^{11}P_{3} &= 11 \times 10 \times 9 = 990 \\ {}^{11}P_{4} &= 990 \times 8 = 7920 \end{aligned} \]

So \(r = 4\). One more factor gives \({}^{11}P_{5} = 7920 \times 7 = 55440\), so 5, 6 and 7 are too large, and 3 gives only 990.

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