nP5 = 120 x nC3
First you have to understand the meaning of the notation. I assume you are solving for n.
nP5 is the number of ways of selecting, in a particular order, 5 distinct objects from a group of n. These are called permutations. The number nP5 equals n(n-1)(n-2)(n-3)(n-4)
nC3 is the number of distinct ways of selecting 3 objects from a group of the same number n, regardless of order. These are called combinations. The number nC3 equals
n(n-1)(n-2)/3!
If
nP5 = 120 nC3, then
n(n-1)(n-2)(n-3)(n-4)
= 120*n(n-1)(n-2)/3! ,
then
(n-3)(n-4) = 20
which requires that
n = 8
nP5/nC3=12
To solve the equation nP5 = 120 x nC3 step-by-step, we will begin by expanding the permutation and combination expressions.
Step 1: Expand nP5 (permutation of n chosen 5):
nP5 = n! / (n-5)!
Step 2: Expand nC3 (combination of n chosen 3):
nC3 = n! / (3! * (n-3)!)
Step 3: Substitute the expanded forms into the equation:
(n! / (n-5)!) = 120 x (n! / (3! * (n-3)!))
Step 4: Simplify the equation by canceling out the common factors, which are the factorials:
(n! / (n-5)!) = 120 x (1 / (3! * (n-3)!))
Step 5: Simplify (3!):
(n! / (n-5)!) = 120 x (1 / (6 * (n-3)!))
Step 6: Simplify 6:
(n! / (n-5)!) = 20 x (1 / ((n-3)!))
Step 7: Multiply through by (n-3)! to remove the denominator:
n! = 20(n-3)!
Step 8: Divide both sides of the equation by (n-3)!:
n!(n-3)! = 20(n-3)!(n-3)!
Step 9: Cancel out the common factors, which are (n-3)! from both sides:
n! = 20(n-3)!
Step 10: Divide both sides of the equation by n!:
1 = 20(n-3)! / n!
So, the equation nP5 = 120 x nC3 can be simplified to 1 = 20(n-3)! / n! in step-by-step form.
To solve the equation nP5 = 120 x nC3, we need to understand what nP5 and nC3 represent.
nP5 stands for permutation of n objects taken 5 at a time. It represents the number of arrangements or orderings of 5 objects out of a set of n objects.
nC3 stands for combination of n objects taken 3 at a time. It represents the number of ways to select 3 objects out of a set of n objects, disregarding the order.
To solve this equation, we need to find the value of n, which satisfies the equation.
First, let's calculate nP5. The formula for permutation is:
nP5 = n! / (n - 5)!
Here, "!" represents the factorial operator, which means multiplying a number by all positive integers less than itself down to 1.
Next, let's calculate nC3. The formula for combination is:
nC3 = n! / (3! * (n - 3)!)
Now, let's substitute these values back into the original equation:
n! / (n - 5)! = 120 * (n! / (3! * (n - 3)!))
To simplify, we can cancel out the n! term on both sides of the equation:
1 / (n - 5)! = 120 / (3! * (n - 3)!)
To further simplify, we can expand the factorials:
1 / (n - 5)! = 120 / (6 * (n - 3) * (n - 4) * (n - 5)!)
Now, we can cross multiply:
(n - 5)! * 120 = 6 * (n - 3) * (n - 4) * (n - 5)!
Next, we can divide both sides by (n - 5)! to isolate the n term:
120 = 6 * (n - 3) * (n - 4)
Lastly, we can simplify further and solve for n:
120 = 6n^2 - 54n + 72
Rearranging the equation:
6n^2 - 54n + 72 - 120 = 0
6n^2 - 54n - 48 = 0
Divide the equation by 6 to simplify:
n^2 - 9n - 8 = 0
Now, we can factorize the quadratic equation:
(n - 8)(n + 1) = 0
This gives us two possible solutions for n:
n - 8 = 0 => n = 8
n + 1 = 0 => n = -1
So, the equation nP5 = 120 x nC3 has two possible solutions for n: 8 and -1.