Permutation Restrictions

[dagr8est]

(n)P(2) = 42
n!/(n-2)! = 42
...

Someone told me that n≥0, but doesn't the second step imply that n ≥ 2? Otherwise you would be taking the factorial of a negative number which is undefined?

 

[stapel]

Are you trying to solve "nP2 = 42" for the value of n? (FYI: Including the instructions can be very helpful.) If so then, yes, the "2" in "n-permute-two" does sort of require that there be at least two objects/people/whatevers to permute.

Note:

. . .$\displaystyle \large{\frac{n!}{(n-2)!}=\frac{1\times 2\times 3\times ...\times (n-3)(n-2)(n-1)(n)}{1\times 2\times 3\times ...\times (n-3)(n-2)}}$

...or, after cancelling duplicate factors, n - n². Set this equal to the given value, and solve the quadratic equation for n.

As for "undefined", that sort of depends on your level of math. In pre-calculus algebra, yeah, I think you can safely assume factorials to apply only to whole numbers, not negatives or other stuff.

Eliz.