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...if we cancel out $n!$ on both sides we get to a complex quadratic which gives a wrong result. But, if we cancel out the $(n-5)!$ and $(n-3)!$ on their respective sides of the equation and then solve the quadratic and use the constraint $n>4$ we arrive at an answer of $n = 10$. Why does the first approach not give the same result? I am completely baffled...is it to do with cancelling out the $n!$ on both sides or is it something else altogether? Please help. Thanks

$nP5 = 42. nP3$, therefore $$\frac{n!}{(n-5)!} = 42 \cdot \frac{n!}{(n-3)!}$$...if we cancel out $n!$ on both the sides, then we get: $$1 = 42(n-5)(n-4)$$ ... the solution of which leads to a quadratic with relatively large numbers and so on and so forth...on the other hand, if we keep the respective sides of the equation separate and solve them, then we get: $$n(n-1)(n-2)(n-3)(n-4)=42. [n(n-1)(n-2)]$$ ...and since $n>4$, $n(n-1)(n-2)$ is not equal to zero, hence dividing both sides by that yields: $$(n-3)(n-4)=42$$ ...which gives an answer of $n=10$

amWhy
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Divesh M
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  • nP5 = 42. nP3, therefore n!/(n-5)! = 42 [n!/(n-3)!]...if we cancel out n! on both the sides, then we get: 1 = 42(n-5)(n-4)...the solution of which leads to a quadratic with relatively large numbers and so on and so forth...on the other hand, if we keep the respective sides of the equation separate and solve them, then we get: n(n-1)(n-2)(n-3)(n-4)=42. [n(n-1)(n-2)]...and since n>4, n(n-1)(n-2) is not equal to zero, hence dividing both sides by that yields: (n-3)(n-4)=42...which gives an answer of n=10 – Divesh M Jun 27 '17 at 15:08
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    please format your question properly so people can more easily understand it and help you better. Here's a mathjax tutorial :) – Dando18 Jun 27 '17 at 15:13
  • Can you tell us your complex quadratic because I am arriving at answer from both ways?? – Atul Mishra Jun 27 '17 at 15:24
  • From your context I infer that "nP5" means "permutations of $n$ things arranged five at a time. I can make the notation a bit more conventional if you confirm this interpretation. – hardmath Jun 27 '17 at 18:55

3 Answers3

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Your mistake was probably assuming that $$\frac{(n-3)!}{(n-5)!}=\frac{1}{(n-4)(n-5)}$$ But notice that $n-5<n-3$ so $(n-3)!=(n-3)(n-4)(n-5)!$ and not $(n-5)!=(n-5)(n-4)(n-3)!$

kingW3
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You wish to solve the equation $$P(n, 5) = 42P(n, 3)$$ for $n > 4$. Since $$P(n, k) = \frac{n!}{(n - k)!}$$ you obtained $$\frac{n!}{(n - 5)!} = 42 \cdot \frac{n!}{(n - 3)!}$$ Canceling $n!$ from each side of the equation yields $$\frac{1}{(n - 5)!} = 42 \cdot \frac{1}{(n - 3)!}$$ Multiplying both sides of the equation by $(n - 3)!$ yields $$\frac{(n - 3)!}{(n - 5)!} = 42$$ Expanding $(n - 3)!$ gives $$\frac{(n - 3)(n - 4)(n - 5)!}{(n - 5)!} = 42$$ Canceling yields $$(n - 3)(n - 4) = 42$$ which is what you obtained using your other method.

N. F. Taussig
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    Thanks a lot Taussig, kimW3 and fleablood! I see where I made the mistake. Thanks again for helping out. Cheers! – Divesh M Jun 29 '17 at 14:19
  • @DiveshM If you only place your thanks under my comment, kimW3 and fleablood will not receive it. Once you have acquired sufficient reputation, you will be able to thank those who provided helpful answers by casting an up vote or accepting what you consider the best answer (assuming you have received one, of course). – N. F. Taussig Jun 29 '17 at 15:15
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$\frac {(n-5)!}{(n-3)!} \ne (n-5)(n-4)$ as $(n-5)! \ne (n-3)!*[(n-4)(n-5)]$

Notice: $(n-5) < (n-3)$ and $(n-3)! = 1*2*3.....*(n-5)*(n-4)*(n-3) = (n-5)!*[(n-4)(n-3)]$ .

And therefore $\frac {(n-5)!}{(n-3)!} = \frac 1{(n-4)(n-3)}$

This will give you the right answer.

fleablood
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