finding $n$ from two permutations.

Question

${^np_4=\ 5(^np_4)}$

I can't find the value of n.

What I have done:

${\frac{n!}{(n-4)!}=\ 5\left(\frac{n!}{(n-4)!}\right)}$

${\frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}=\ 5\frac{n(n-1)(n-2)(n-3)(n-4)!}{(n-4)!}}$

${n(n-1)(n-2)(n-3)=\ 5n(n-1)(n-2)(n-3)}$

What do I do next or is this question wrong?

Edit:

Guess the question is wrong from all of your answers. it was asked on a exam.

Answer 1

This question is (almost) certainly in error. What you ultimately have is an equation of the form

$$x = 5x$$

where $x = P(n,4)$. This would only hold if $x=0$, which $P(n,4)$ obviously never is*, so this amounts to the statement $1=5$ for all intents and purposes.

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* Obviously, only for $n \ge 4$. If we interpret it for integers $0<n<4$, $P(n,4)$ amounts to "how many ways can we make arrangements of $4$ items from a set of $n<4$ items?" In that sense, $P(n,4) = 0$ for $0<n<4$, clearly: you can't even get the four items to begin with.

So in that sense you could say there does exist a solution: that being $0<n<4$ (and also $n$ an integer but that has to do with stuff involving the gamma function, which generalizes the factorial to non-integer values, which is probably beyond the scope of this post).

Granted this also raises to me the question of how to handle negative or non-integer $n$, now that drhab noted this slight oversight in the comments. But I suppose that's an issue for me to find on my own. :p

Anyhow I maintain it's still probably poor design on the part of the question author, but if there is indeed a solution, you have four: $n=0,1,2,3.$

Answer 2

The question is wrong 100%. nPr and nCr have fixed NON ZERO values and there is no reason why the question statement is logical. Most probably a misprint.

Answer 3

If $^nP_4$ stands for the "number of ordered $4$-tuples of distinct elements of $\{1,\dots,n\}$" then: $$^nP_4=\frac1{4!}n(n-1)(n-2)(n-3)$$

So that $^nP_4=0$ if $n\in\{0,1,2,3\}$.

(and indeed: $4$-tuples of distinct elements of e.g. $\{1,\dots,3\}$ do not exist)

Also the following definition can be practicized: $$^nP_4=4!\binom{n}{4}$$ and there is a (very useful) convention that $\binom{n}{k}$ is defined for nonnegative integer $n$ and integer $k$ under $\binom{n}{k}=0\iff k\notin\{0,1,\dots,n\}$

Answer 4

let $x=$$\frac{n!}{(n-4)!}$

$$x-5x=0 \rightarrow x=0$$ it is impossible because the $\frac{n!}{(n-4)!}\geq 4!$