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I am trying to understand how simplified that equation:$$\frac{(x-2)!}{(x-4)!}-46=\frac{x!}{(x-1)!}$$
In my book it was solved like this:
but I really don't understand how it happened? Where did $(x-3)(x-4)(x-5)$ come from in first fraction? It looks like I missed something fundamental about factorials...
$\endgroup$ 21 Answer
$\begingroup$$$\frac{(x-2)!}{(x-4)!}-46=\frac{x!}{(x-1)!}\iff \frac{(x-2)!}{(x-4)!}-\frac{x!}{(x-1)!}-46=0$$$$\frac{(x-2)!}{(x-4)!}-\frac{x!}{(x-1)!}-46= \frac{(x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)}\cdot \frac{(x-2)!}{(x-4)!}-\frac{x!}{(x-1)!}-46= $$$$=\frac{[(x-2)(x-3)]\cdot(x-1)!}{(x-1)!}-\frac{x\cdot(x-1)!}{(x-1)!}-\frac{46(x-1)!}{(x-1)!}=(x-2)(x-3)-x-46 $$
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