Question

If $5!=5\times 4\times 3\times 2\times 1$
$8!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$ then what is $-8!=$? and $-5!=$? What is the elongation of -ve factorial?

Answer 1

The factorials of negative integers have no defined meaning.
However, we can generalize the factorial function to the Gamma function
$\Gamma \left(x\right)={\int }_0^{\infty }{t}^{x-1}{e}^{-t}dt$
by noting that, if we integrate the above by parts, we obtain
$\Gamma \left(x\right)=(x-1)\Gamma (x-1)$
and since $\Gamma \left(1\right)=1$ (try it!), we obtain that
$\Gamma \left(n\right)=(n-1)!$ for any positive integer n.
The Gamma function can also be extended to complex numbers by replacing the real number x in the above definition by a complex number z.
Note that the Gamma function is undefined for nonpositive integers, though it is defined for every other number, including complex numbers as said already.
To summarize: the factorial function n! is only defined for non negative values of n. If you wish a more general form of the factorial function, you'll need to use the Gamma function $\Gamma \left(z\right)$, which is defined for every complex value of z except z = 0, -1, -2, -3 ....
1) (15 - n)!
2) (n - 7)!
Ex: We have to prove -
$\frac{(15-n)!}{(12-n)!}=5!$
$\Rightarrow \frac{(15-n)(14-n)(13-n)(12-n)!}{(12-n)!}=\mathrm{5.4.3.2.1}$
$\Rightarrow (15-n)(14-n)(13-n)=120$
As L.H.S is product of three consecutive natural numbers, hence we should try to break R.H.S too in product of three consecutive integers.
120 = 6 . 5 . 4
Hence
(15 - n) (14 - n) (13 - n) = 6 .5 .4
On L.H.S largest number is 15 - n and on R.H.S largest number is 6
$\Rightarrow$ 15 - n = 6
$\Rightarrow$ n = 15 - 6
$\Rightarrow$ n = 9