Question

(n+2)!/(4!(n-2)!) = 6n!/(2!(n-2)!)

can someone check my solution? i can't find my mistake. the value of n should be 0, 1, 7

2(n-2)!(n+2)!=6n!4!(n-2)!
2(n+2)!=6n!4!
(n+2)(n+1)n!=3n!4!
(n+2)(n+1)=72
which doesn't equal to n=0, 1, 7 :(

Answer 1

n=7 works

Answer 2

from (n+1)(n+2)=72 you get \(0=n^2+3n-70=(n-7)(n+10)\)
So n=10 since we want n to be positive integer

Answer 3

\[2(n+2)!=6n!4!\]
\[(n+2)(n+1)n!=72n!\]
\[(n+2)(n+1)=72\]

Answer 4

I mean n=7

Answer 5

answer is supposed to be n=0, 1, 7.... there's no n=10 :(

Answer 6

so what is your definition for n! where n is negative ? because when we plug in n=0 we have the (-2)! expression

Answer 7

normally n! = 1 if n <= 0

Answer 8

@Hero @ParthKohli @Yahoo! @dumbcow @SmoothMath @KatrinaKaif @TheViper ? anyone? :D please! thank you!

Answer 9

idk i keep getting

(n+2)(n+1)(n)(n-1) = 72n(n-1)

every time i do it

what are you looking for?? the solutions you provided are all vaild

Answer 10

Okay, you've got \[(n+2)(n+1)n!/(4!(n-2)!) = 6n!/(2!(n-2)!)\]

Now we can write \[(n-2)!\] as \[n!/{n(n-1)}\] which gives us
\[(n+2)(n+1)n!/(4!(n!/{n(n-1)}))) = 6n!/(2!(n!/{n(n-1)}))\]

Because that gives you n(n-1)*<stuff> on both sides, you'll have solutions for n = 0 and (n-1)=0 because the equation will reduce to 0*<stuff> = 0*<stuff>.

Answer 11

(n+2)!/(4!(n-2)!) = 6n!/(2!(n-2)!)
(n+2)(n+1)n! /4! =6 n!/2!
(n+2)(n+1) = 12*6
Now, just solve for n

Answer 12

when n=0 or 1
(n-2)! is not definied

Answer 13

@whpalmer4 how come (n-2)! is equal to n!/n(n-1) ???

Answer 14

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

Answer 15

@whpalmer4 and @binarymimic, thank you! but i'd also want to know where i made my mistake and why we're not reaching the same answers.

Answer 16

see the 10th post in this thread where I show how you get the n = 0 and n = 1 solutions

Answer 17

\[\frac{ (n+2)! }{ 4!(n-2)! } = \frac{ 6n! }{ 2!(n-2)! }\]

\[\frac{ (n+2)(n+1)(n)(n-1)(n-2)! }{ 24(n-2)! } = \frac{ 3n(n-1)(n-2)! }{ (n-2)! }\]

\[\frac{ (n+2)(n+1)(n)(n-1) }{ 24 } = 3(n)(n-1)\]

\[(n+2)(n+1)(n)(n-1) = 72(n)(n-1)\]

from here you can see that if n = 0 or n = 1 then both sides reduce to 0
so we can factor out n(n-1)

\[(n+2)(n+1) = 72\]

you would have to exclude any solutions that are not part of the original domain though

Answer 18

so i am not sure why 0 and 1 are acceptable answers since they are not part of the original domain. only n = 7 is part of the original domain

Answer 19

n=0 cannot be a solution. Besides the definition of n! for n=0,1,2... there is the generalisation of it: the Gamma function. n! = Gamma(n+1). The Gamma function extends the factorial for non-integer values.
Negative values are also allowed, except for -1, -2, -3, .... The graph of the Gamma function has vertical asymptotes for -1, -2, -3, ...
Putting n=0 in the equation leads to (-2)! which is undefined.

Solve the equation:\[\frac{ (n+2)! }{ 4!(n-2)! }=\frac{ 6n! }{ 2!(n-2)! } \Leftrightarrow \frac{ (n+2)! }{ 4! }=\frac{ 6n! }{ 2 }=\frac{3n!}{1}\]So\[(n+2)!=3 \cdot 4!n!=72n! \Leftrightarrow n!(n+1)(n+2)=72n!\]Divide by n! (which is never 0):\[(n+1)(n+2)=72 \Leftrightarrow n^2+3n+3-72=0\]\[n^2+3n-70=0 \Leftrightarrow (n+10)(n-7)=0\]This has solutions -10 and 7.
-10 is not valid because it gives (-12)! in the original equation.

Conclusion: 7 is the only solution.

Just to be safe: try all alleged solutions, by substituting them in the original equation:

n=0:\[\frac{ 2 }{ 24(-2)! } \neq \frac{ 6 }{ 2(-2)! }\]
n=1:\[\frac{ 6 }{ 24(-1)! } \neq \frac{ 6 }{ 2(-1)! }\]
n=7:\[\frac{ 9! }{ 4! \cdot 5! }=\frac{ 6 \cdot7! }{ 2 \cdot 5! } \Leftrightarrow \frac{ 9! }{ 4! }=\frac{ 6 \cdot 7! }{ 2 } \Leftrightarrow\]\[ 5 \cdot 6\cdot 7\cdot 8\cdot 9 =3\cdot4\cdot5\cdot6\cdot7\cdot6 \Leftrightarrow 8 \cdot 9 = 3 \cdot 4 \cdot 6 \Leftrightarrow 72=72\]

This confirms: 7 is the only solution.