Solve for n:
(n-2)!/(n-3)!=15
and
(n-2)!/(n-1)=12
I'll solve the first one, and the second one can be solved similarly.
(n-2)!/(n-3)!
=(n-2)(n-3)!/(n-3!)
=n-2
therefore n-2=15 => n=17
Check:
(17-2)!/(17-3)!=15!/14!=15*14*...*1/(14*13*..*1)
=15 checked ok.
To solve these equations for n, we can start by simplifying the expressions on the left-hand sides of the equations.
Equation 1: (n-2)!/(n-3)!=15
We'll start by simplifying the factorial expressions. The factorial (n-2)! means we need to multiply all the integers from (n-2) down to 1. Similarly, (n-3)! means we need to multiply all the integers from (n-3) down to 1.
Let's assume n is a positive integer greater than 3, otherwise, the equations would not be defined.
(n-2)! = (n-2)(n-3)!
(n-3)!/(n-3)! = 15
Now, we can cancel out the (n-3)! terms:
1 = (n-2)
Simplifying further:
n - 2 = 1
n = 3
Therefore, the solution to the first equation, (n-2)!/(n-3)! = 15, is n = 3.
Equation 2: (n-2)!/(n-1) = 12
Using the same logic:
(n-2)!/(n-1) = 12
Again, let's simplify the factorial expression:
(n-2)! = (n-2)(n-3)!
(n-2)(n-3)!/(n-1) = 12
Now, we can cancel out the (n-3)! terms:
(n-2)/(n-1) = 12
Multiplying both sides by (n-1) gives:
(n-2) = 12(n-1)
Expanding and simplifying:
n - 2 = 12n - 12
Combining like terms:
11n = 10
n = 10/11
Therefore, the solution to the second equation, (n-2)!/(n-1) = 12, is n = 10/11.
In summary, the solutions to the given equations are n = 3 and n = 10/11.