How do u solve for n Permutations

[gillgill]

how do u solve for n?
nP3=720...n!/(n-3)!=720...do u just start cancelling?
that will be n(n-1)(n-2)=720...but its such a big number...is there another easier way to do it?

 

[xanthym]

how do u solve for n?
nP3=720...n!/(n-3)!=720...do u just start cancelling?
that will be n(n-1)(n-2)=720...but its such a big number...is there another easier way to do it?

SOLUTION HINTS:
Factor 720 to help find the solution:
720 = (2⁴)*(3²)*(5) = n*(n - 1)*(n - 2) = (?)*(?)*(?)
(Hint: Try 2³)


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[MiniTank]

how do u solve for n?
nP3=720...n!/(n-3)!=720...do u just start cancelling?
that will be n(n-1)(n-2)=720...but its such a big number...is there another easier way to do it?

by cancelling do you mean use the factor theorem? because i would expand and move the 720 to the left side and factor it out