MediumMathematicsHigh School (11–12)Algebra › Permutations and CombinationsNumerical

For some n and r, ⁿPr = 336 and ⁿCr = 56. Find n and r.

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For some $n$ and $r$ , $^{n}P_{r} = 336$ and $^{n}C_{r} = 56$ . Find $n$ and $r$ .

Need a reminder? The key ideaShow

A combination is a selection in which order does not matter. $^{n}C_{r} = \dfrac{n!}{r!\,(n - r)!} = \dfrac{n(n - 1)\cdots(n - r + 1)}{r!}$ , and $^{n}P_{r} = {}^{n}C_{r} \times r!$ . Also $^{n}C_{0} = {}^{n}C_{n} = 1$ and $^{n}C_{2} = \dfrac{n(n - 1)}{2}$ .

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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← Lalit owns 12 different books. He will take 5 of them on a holiday.If ⁿC₁₂ = ⁿC₈, find n and then ⁿC₁₇. →

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For some n and r, ⁿPr = 336 and ⁿCr = 56. Find n and r.

The worked solution

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