n!/(n-2)!=90
n=10
how did u get the n=10
rewrite it as ...
n!/(n-2)!=90
[n(n-1)(n-2)(n-3)..(2)(1)]/(n-2)(n-3)..(2)(1)] = 90
n(n-1) = 90
n^2 - n = 90 = 0
(n-10)(n+9)=0
n=10 or n= -9 , but n > 0
so n=10
To solve the equation n!/(n-2)! = 90, we can simplify the expression by canceling out the (n-2)! terms:
n!/(n-2)! = 90
n * (n-1) * (n-2)! / (n-2)! = 90
n * (n - 1) = 90
Now we have a quadratic equation. Simplify and rearrange it to solve for n:
n^2 - n - 90 = 0
Factoring the quadratic equation gives:
(n - 10)(n + 9) = 0
Setting each factor equal to zero, we get n = 10 and n = -9.
However, we cannot have a negative factorial value, so we can disregard n = -9. Therefore, the solution to the equation n!/(n-2)! = 90 is n = 10.