Permutation Restrictions
- Thread starter dagr8est
- Start date
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- Feb 4, 2004
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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 - n<sup>2</sup>. 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.
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 - n<sup>2</sup>. 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.