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

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.

Hadaf
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  • either the question is wrong or you copied wrong. how can nP4 = 5nP4?? that's like 1=5 – Saketh Malyala May 01 '19 at 08:42
  • If $^np_4$ stands for the number of ordered $4$-tuples of distinct elements of ${1,\dots,n}$ then evidently $^np_4=0$ if $n\in{0,1,2,3}$ – drhab May 01 '19 at 09:06

4 Answers4

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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.

* 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: $n=0,1,2,3.$

PrincessEev
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  • If e.g. $P(3,4)$ is defined then evidently it should be defined as $P(3,4)=0$, so implicitly you are saying that $P(3,4)$ is not defined. Why not? – drhab May 01 '19 at 09:07
  • It actually hadn't crossed my mind, if I'm being honest. Which is kind of dumb on my part since I'm aware of such generalizations of binomial coefficients. >_> EDIT: And I also completely misinterpreted your point. Oops. – PrincessEev May 01 '19 at 09:09
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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 $\binom{n}{k}=0\iff k\notin\{0,1,\dots,n\}$

drhab
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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.

  • What side are you on? Is $3P4=0=5\times0=5\times3P4$ true or not? – drhab May 01 '19 at 09:18
  • Isn't nPr always non zero??? –  May 01 '19 at 09:19
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    That's what I was thinking, @PranavGupta53535, but you probably made the oversight I did. How many ways can you take $4$ distinct objects from $3$ and arrange them? One answer, $P(3,4)$. The other, obviously, $0$, since $3<4$. – PrincessEev May 01 '19 at 09:22
  • Does nPr have a formula your n<r? Isn't it undefined? –  May 01 '19 at 09:26
  • Logically, your point make sense though... –  May 01 '19 at 09:27
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    There are two ways we often end up defining $P(n,r)$ - as a product or "falling factorial", and probably the formula you're most familiar with. They are respectively

    $$P(n,r) = n(n-1)(n-2)\cdots(n-r+1) = \frac{n!}{(n-r)!}$$

    We use the first formula in defining a "generalized" binomial coefficient for $n\in\Bbb R$, so I imagine we can do the same here. And then notice in that product, if $n<r$, we eventually have a factor of $0$. Thus, getting $0$.

     – PrincessEev May 01 '19 at 09:40
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    Well, $0<n<r$ to be specific. For some reason Wolfram wants to have $n<0$ undefined even though this convention would let us define such, which raises a question in and of itself. – PrincessEev May 01 '19 at 09:42
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let $x=$$\frac{n!}{(n-4)!}$

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

E.H.E
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