To solve these equations algebraically for n using factorial notation, we need to manipulate the equations to isolate the factorials on one side of the equation.
a) n!/3!(n-2)! = 12
We can start by simplifying the factorials. Since we have n! in the numerator and 3!(n-2)! in the denominator, we can rewrite them as:
n! = 3!(n-2)! * 12
Now, let's expand the factorials:
n * (n-1) * (n-2)! = 6 * (n-2)! * 12
The (n-2)! terms cancel out from both sides, leading to:
n * (n-1) = 6 * 12
n * (n-1) = 72
To solve this equation, we can use the quadratic formula or factor the right-hand side of the equation. In this case, let's factor:
(n-9)(n+8) = 0
Setting each factor equal to zero:
n-9 = 0 or n+8 = 0
n = 9 or n = -8
So the solutions to the equation are n = 9 and n = -8.
b) 8Pn = 6720
The notation 8Pn represents the permutation of 8 objects taken n at a time. It is calculated as:
8Pn = 8! / (8-n)!
Setting it equal to 6720, we get:
8! / (8-n)! = 6720
Now, let's simplify the factorials:
8 * 7 * 6! / (8-n)! = 6720
Notice that 6! can be further simplified to 6 * 5 * 4 * 3 * 2 * 1. So we can rewrite the equation as:
8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / (8-n)! = 6720
To solve this equation, we need to find the value of n for which the left-hand side equals 6720. We can start by simplifying the numerator:
(8-n)! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 / 6720
Now, we can cancel out common factors:
(8-n)! = 4 * 6 * 5
(8-n)! = 120
Since 120 is the factorial of 5 (5!), we have:
8 - n = 5
Solving for n:
n = 8 - 5
n = 3
Therefore, the solution to the equation is n = 3.