Solve the following equations algebraically for n. Show your working using factorial notation.
a) n!/3!(n-2)! = 12
b) 8Pn = 6720
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.
a) To solve the equation n!/3!(n-2)! = 12 algebraically using factorial notation, we can start by simplifying the equation.
First, let's rewrite the factorials in expanded form:
n! = n * (n-1) * (n-2) * ... * 3 * 2 * 1
3! = 3 * 2 * 1
(n-2)! = (n-2) * (n-3) * (n-4) * ... * 3 * 2 * 1
Now we can substitute these values back into the equation:
(n * (n-1) * (n-2) * ... * 3 * 2 * 1) / (3 * 2 * 1) * ((n-2) * (n-3) * (n-4) * ... * 3 * 2 * 1) = 12
Cancel out common terms:
n * (n-1) / (n-2) * (n-3) * (n-4) * ... * 3 * 2 * 1 = 12
Next, simplify the left side:
(n * (n-1)) / (n-2)! = 12
Multiply both sides by (n-2)!:
n * (n-1) = 12 * (n-2)!
Expand n * (n-1):
n^2 - n = 12 * (n-2)!
Rearrange the equation:
n^2 - n - 12 * (n-2)! = 0
Now, let's consider the possible values of (n-2)!. Since the factorial of any positive integer is always positive, we know that (n-2)! is greater than 0. Therefore, we can divide through the equation by (n-2)!:
(n^2 - n) / (n-2)! - 12 = 0
Simplify further:
(n-1)(n) / (n-2)! - 12 = 0
Multiply both sides by (n-2)!:
(n-1)(n) - 12(n-2)! = 0
Expand (n-1)(n):
n^2 - n - 12(n-2)! = 0
We can solve this quadratic equation for n by factoring or using the quadratic formula. However, since n is not specified, we cannot determine its exact value without additional information or simplification.
b) To solve the equation 8Pn = 6720 algebraically using factorial notation, we can start by simplifying the equation.
The permutation formula is given by:
nPn = n!
Therefore, we can rewrite the equation as:
nPn = 6720
Substituting the values using the factorial definition:
8!/ (8-n)! = 6720
Expand 8!:
8*7*6*5*4*3*2*1 / (8-n)! = 6720
Now we can simplify:
40320 / (8-n)! = 6720
Multiply both sides by (8-n)!:
40320 = 6720 * (8-n)!
Divide both sides by 6720:
6 = (8-n)!
Now we need to find the factorial of (8-n)! that equals 6. By inspection, we can see that (8-n)! = 3. Therefore, the equation simplifies to:
8 - n = 3
Subtract 8 from both sides:
-n = 3 - 8
-n = -5
Multiply by -1:
n = 5
The solution to the equation 8Pn = 6720 is n = 5.