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I have a statement that says:

Determine the value of x, so that it is fulfilled:

$$\binom{x}{2} = 10$$

Ok, i try this:

$$\begin{align*} \frac{x!}{(x - 2)! \cdot 2!} &= 10 \\ x! &= 10(2x -4) \\ x! &= 20x - 40 \\ x! - 20x &= 40 \end{align*}$$

I was here, and I did not know what else to do.

How should I solve it? What is my error?

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    $\begingroup$ $(x-2)!\not=(x-2)$ $\endgroup$
    – Sorfosh
    Mar 26, 2018 at 23:42

5 Answers 5

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Hint: $\frac{x!}{(x-2)!}=x(x-1)$

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Note that

$$\frac{x!}{(x - 2)! \cdot 2!} = 10\iff\frac{x(x-1)(x-2)!}{(x - 2)! \cdot 2!} = 10$$

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  • $\begingroup$ How you get $x(x -1)(x-2)!$ $\endgroup$
    – ESCM
    Mar 27, 2018 at 0:23
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    $\begingroup$ By definition $x!=x(x-1)(x-2)(x-3)...2\cdot 1=x(x-1)(x-2)!$ $\endgroup$
    – user
    Mar 27, 2018 at 0:25
  • $\begingroup$ Ah, its so easy now, thanks gimusi $\endgroup$
    – ESCM
    Mar 27, 2018 at 0:58
  • $\begingroup$ @Mattiu You are welcome! Bye $\endgroup$
    – user
    Mar 27, 2018 at 5:53
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There are several good answers here, but I don't think that any of them have directly attempted to address the error you made (though it is pointed out in the comment). There are actually two critiques that I would make regarding your argument, as summarized below:

First, you write \begin{align*} \frac{x!}{(x - 2)! \cdot 2!} &= 10 \\ x! &= 10(2x -4) \\ x! &= 20x - 40 \\ x! - 20x &= 40. \end{align*} I would argue that this is unclear—were I grading your work, I almost certainly would not give full marks to such an answer, even if it were otherwise correct. You have written down a series of statements, but have giving no indication about how they relate to each other. If your intention is to say that one line follows from the next, then you should indicate that in some way, for example: \begin{align*} \frac{x!}{(x - 2)! \cdot 2!} = 10 &\stackrel{(*)}{\implies} x! = 10(2x -4) \\ &\implies x! = 20x - 40 \\ &\implies x! - 20x = 40 \end{align*}

Second, there is a significant error in your computation, which I have highlighted with an (*). You appear to have concluded that $(x-2)!\cdot 2! = (2x-4)$, then attempted to clear the denominator by multiplying both sides of the equation by this term. However, this computation is incorrect. You might be better off noting that the term $$ \frac{x!}{(x-2)!} $$ simplifies rather nicely. Indeed, if you write it out in painful and pedantic detail, you get $$ \frac{x!}{(x-2)!\cdot 2!} = \frac{x(x-1)\color{red}{(x-2)(x-3)\dotsb(3)(2)(1)}}{\color{red}{(x-2)(x-3)\dotsb(3)(2)(1)}\cdot 2} = \frac{x(x-1)}{2}. $$ Therefore \begin{align} 10 = \frac{x!}{(x-2)!\cdot 2!} = \frac{x(x-1)}{2} &\implies 20 = x(x-1) = x^2-x \\ &\implies 0 = x^2 - x - 20 = (x-5)(x+4) \\ &\implies x=5 \quad\text{or}\quad x=-4. \end{align} From the start, we know that $x > 0$, therefore $x=5$.

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From $$\frac{x!}{(x - 2)! \cdot 2!} = 10$$ we get $$\frac{x(x-1)(x-2)!}{(x-2)!\cdot 2}=10$$ Then $$\frac{x(x-1)}2=10\quad\text{so}\quad x(x-1)=20$$

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  • $\begingroup$ How you get $x(x -1)(x-2)!$ $\endgroup$
    – ESCM
    Mar 27, 2018 at 0:23
  • $\begingroup$ @Mattiu Expand the first couple of terms of $x!$. $\endgroup$
    – amd
    Mar 27, 2018 at 2:58
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You can be lazy and just check if any solution works (it might have been the case there was no solution.... in general there shouldn't always be a solution).

By direct computation $\binom 2 2=1, \binom 3 2= 3, \binom 4 2 = 6, and \binom 5 2 =10 $

This seems like a natural approach since for fixed $n$, $f(m)=\binom m n$ is an increasing function, so that if $f(m)< k$ and $f(m+1)>k$ then there is no solution to be found. But there is a solution... $f(5)=10$.

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