Question

Could someone show me how to solve this?
(n!)/(n-3)!=5000

Answer 1

\[\frac{n!}{(n-3)!}=\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}=n(n-1)(n-2)\]

Answer 2

Yes well I need my result to be a whole number..... Like I need to solve for n

Answer 3

Could you show me how to get be the number result for n?

Answer 4

@SithsAndGiggles

Answer 5

Does anybody know?? ?

Answer 6

@wolf1728 @campbell_st

Answer 7

(n!)/(n-3)!=5000 can be rewritten
(n * n-1) / (n-3)! = 5000
I'm working on it

Answer 8

well the way I see it... you can remove common factors from the numerator and denominator and are left with the equation

n(n-1)(n -2) = 5000

or

\[n^3 - 3n^2 + 2n -5000 = 0\]

so its a process of solving for n

or finding 3 consecutive numbers that multiply to 5000

Answer 9

What do you mean @campbell_st ?

Answer 10

ok... n! = n x (n -1) x (n -2) x (n -3) x(n -4) ....... x 2 x 1

(n -3)! = (n -3) x (n -4) x (n -5) x.....2 x 1

so if you eliminate common factors from the numerator and denominator

you get

n x (x -1) x (n -2) = 5000

Answer 11

So what would that result to? Like what whole number would n equal. That is what i am trying to solve

Answer 12

@campbell_st

Answer 13

the problem is this doesn't give n as a integer...

Answer 14

I don't know..lol I am rusty in this @campbell_st

Answer 15

it means n = 18.1193

Answer 16

hmm okay

Answer 17

are you sure its not an inequation...?

Answer 18

I give up lol

Answer 19

well its not something like

largest n so that

\[\frac{n!}{(n -3)!} < 5000\]

Answer 20

Yes, I also get a non-interger solution too
n=18.119

Answer 21

tough question the way its written.... I was expecting n was a positive integer

Answer 22

BaileyC - are you absolutely sure of the way the question is written?

Answer 23

u r left with n(n-1)(n-2)=5000
n[n^2-2n-n+2]=5000
n[n^2-3n-2]=5000