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Worked Examples · Example 4

Q.Find nn if nP3=60^{n}P_{3} = 60.

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By definition nP3=n!(n−3)!=n(n−1)(n−2)^{n}P_{3} = \dfrac{n!}{(n-3)!} = n(n-1)(n-2) — the product of three consecutive integers ending at nn. Set it equal to 6060:

n(n−1)(n−2)=60.n(n-1)(n-2) = 60.

Recognise 60=5×4×360 = 5 \times 4 \times 3, three consecutive integers, so n=5n = 5. …

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