How do i solve algebraically for n. 9n=3(nP2)

Answer 1

n. 9n=3(nP2)
=>9n^2=3(n!/(n-2)!)
=>9n^2=3(n*(n-1)(n-2)!/(n-2)!)
=>9n^2=3(n*(n-1))
=>9n^2=3(n^2-n)
=>9n^2=3n^2-3n
=>6n^2-3n=0
=>3n(n-2)=0
=>n=0 and n=2

Answer 2

Did you understand harley?

Answer 3

Yes, i just didn't know how to start with the 9n, thank you!

Answer 4

oops,I made a mistake

6n^2+3n=0
=> 3n(2n+1)=0
=>n= 0 and n= -1/2

Answer 5

The mistake was in starting with 9n and the saying 9n^2.

9n=3(nP2)
=>9n=3(n*(n-1))
=>9n=3n^2-3n
=>3n^2-12n = 0
=> 3n(n-4) = 0
=> n = 0,4