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I have the following equation: $${{x}\choose{3}}=10$$ I want to solve it for $x$ so I wrote: $${{x}\choose{3}}=\frac{x!}{3!(x-3)!}=10$$ which follows: $$\frac{x!}{(x-3)!}=60$$ I wonder how to proceed from here and solve for $x$? As $(x-3)!\neq x!-3!$.

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  • $\begingroup$ Think how to further simplify the left-hand side with the division. Plug in the definition of the factorial. $\endgroup$ – Matti P. May 11 '18 at 10:44
  • $\begingroup$ Would you accept $-1 \pm\sqrt{-11}$? $\endgroup$ – Henry May 11 '18 at 15:55
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Hint: Simplifying $$\binom{x}{3}=\frac{x(x-1)(x-2)}{1\cdot 2\cdot3}$$ we get $$\frac{x^3-3x^2+2x}{6}=10$$

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$$x!=1\cdot2\cdot3\cdot(...)\cdot(x-3)\cdot(x-2)\cdot(x-1)\cdot x$$ $$3!(x-3)!=6\cdot1\cdot2\cdot3\cdot(...)\cdot(x-3)$$ Hence $$\frac{x!}{3!(x-3)!}=\frac{x(x-2)(x-1)}{6}=10$$ and solve from there.

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$$\frac{x!}{(x-3)!} = x(x-1)(x-2)$$
as $x(x-1)(x-2)=60$ which is a cubic equation it can easily be solved
giving $x=5$

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Observe that $$\binom{x}{3} = \frac{(x)(x-1)(x-2)}{6}$$

$$ \frac{(x)(x-1)(x-2)}{6} = 10 \implies (x)(x-1)(x-2)-60 = 0$$ $$ x^3 -3x^2+2x-60 = 0$$

You can factorise the above cubic equation to finish off

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