While solving for $nP5 = 42 \cdot nP3$, $n > 4$ ...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 $n=10$

$\frac {(n-5)!}{(n-3)!} \ne (n-5)(n-4)$ 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.
