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So here's my attempt : $${(n-1)(n-2)(n-3)! \over (n-3)!} = 30$$

$${(n-1)(n-2)} = 30$$

then $$n-1 = 30$$ or $$n-2 = 30$$ then $$n = 31$$ or $$n = 32$$

but my textbook says

$$(n-1)(n-2) = 6 * 5$$

then $$(n-1) = 6$$ then $$n = 7$$

which does make any sense to me this should have been like any other factorize , can someone please explain the steps of the textbook to me , thanks in advance

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  • $\begingroup$ Just a little note: AB=0 means either either A=0 or B=0. In your attempt you are saying 30*29=30 $\endgroup$ – randomgirl Apr 20 at 15:32
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    $\begingroup$ .. but $AB = 30$ does NOT imply that $A = 30$ or $B=30$. $\endgroup$ – NickD Apr 20 at 15:33
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$$\require{cancel} \frac{(n-1)!}{(n-3)!}=30\\ \frac{\cancel{(n-3)!}(n-2)(n-1)}{\cancel{(n-3)!}}=30\\ (n-2)(n-1)=30\\ n^2-3n+2=30\\ n^2-3n-28=0 $$

All you have to do now is solve that quadratic equation in $n$.

$$ n=\frac{3\pm\sqrt{9+4\cdot28}}{2\cdot1}=\frac{3\pm11}{2}\implies\\ n=7,\ n=-4 $$

$n=-4$ should be discarded as an extraneous solution because factorials of negative numbers are not defined (if you plug it into the original expression, it will give you factorials of negative numbers which cannot be calculated).

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You started off well, the equation $(n-1)(n-2)=30$ is correctly derived. Then you make a case distinction which is completely incorrect.

Instead, simply solve the quadratic equation $n^2-3n+2=30$.

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You've got the first bit spot on, $(n-1)(n-2)=30$

Now, you say that either $(n-1)=30$ or $(n-2)=30$. This would be true only if the other was equal to $1$ at the same time. If $n-1=30$, $n-2=29$ clearly, so this is not met, and likewise if $(n-2)=30$ then $(n-1)=31$ this is not met either.

Instead, expand out $(n-1)(n-2)=n^2-3n+2=30 \to n^2-3n-28=0$

I know $4+(-7)=-3\text{ and } 4\cdot(-7)=-28$ so $(n+4)(n-7)=0\implies n=7, -4$

It can't be negative thus $n=7$

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$$(n-1)(n-2)=30 \ \lnot\implies (n-1)=30 \lor (n-2)=30 $$


$$\begin{aligned}(n-1)(n-2)&=30\\ n^2-3n+2&=30\\ n^2-3n-28&=0\end{aligned}$$

Your textbook has a good solution wherein we're supposed to identify that $30=(7-1)(7-2)$. This can also be achieved by solving the quadratic equation so formed which factorises nicely as $(n-7)(n+4)$ and gives $n=7$ as the only possible solution.

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You started correctly but then in the $2^{nd}$ step you did a principal mistake in your work.

If $(n-1)(n-2)=30$ this doesn't imply that $n-1=30$ or $n-2=30$

You do this ONLY in the case of $0$. I mean if $(n-1)(n-2)=0 \implies n-1=0$ or $n-2=0$

But in your exercise you have $(n-1)(n-2)=30$ so $(n-1)$ and $(n-2)$ are 2 consecutive natural numbers (i.e. they differ only by $1$) and, their product is $30$ (i.e. $(n-1)×(n-2)=30$)

So you must think of 2 numbers such that their product is $30$ (these are the divisors of $30$)

So we have, $1×30=30$ or $2×15=30$ or $3×10=30$ or $5×6=30$

Now you can easily notice that the only pair of numbers that differ only by $1$ (or are consecutive) are $5$ and $6$

And ofcourse $5$ is less than $6$ so $n-1=6$ and $n-2=5$

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