How would i solve for n

nP5 = 14(nP4)

nP5 = n!/(n-5)!

your equation:
n!/(n-5)! = 14n!/(n-4)!
n!/(n-5)! = 14n!/(n-4)(n-5)!
1 = 14/(n-4)
n-4 = 14
n = 18

check:
my calculator says P(18,5) = 1028160
and P(18,4) = 73440
and 14(73440) = 1028160

Why did the number go to the therapist? Because it had a recurring complex!

To solve for "n" in the equation nP5 = 14(nP4), let's break it down:

nP5 represents the number of permutations of "n" objects taken 5 at a time.
nP4 represents the number of permutations of "n" objects taken 4 at a time.

To solve the equation, we need to find the value of "n" that makes the equation true.

Now, let's expand the equation:
nP5 = 14(nP4)

The number of permutations can be calculated using the factorial function:
nP5 = n! / (n-5)!

Expanding the right side of the equation:
14(nP4) = 14 * n! / (n-4)!

Now, equating the two sides of the equation, we have:
n! / (n-5)! = 14 * n! / (n-4)!

To get rid of the factorials, we can cancel them out:
(n! / (n-5)!) = (14 * n! / (n-4)!)

Cancelling out the factorials gives us:
1 / (n-5)! = 14 / (n-4)!

To solve for "n", we can cross-multiply:
14 * (n-5)! = (n-4)!

Dividing both sides by 14:
(n-5)! = (n-4)! / 14

Now, we have an equation without factorials. To solve it, we'll need to approach it symbolically or use a numerical method.

To solve for n, you need to simplify the equation by expanding the permutations (nP5 and nP4) and then solve for n.

The formula for permutations is given by:

nPk = n! / (n - k)!

Where n! is the factorial of n, denoting the product of all positive integers less than or equal to n.

Let's expand the permutations:

nP5 = n! / (n - 5)!
nP4 = n! / (n - 4)!

Substituting these into the equation:

n! / (n - 5)! = 14 * (n! / (n - 4)!)

Now, to simplify further, we can cancel out the factorials:

(n! / (n - 5)!) / (n! / (n - 4)!) = 14

Dividing fractions is the same as multiplying the first fraction by the reciprocal of the second fraction:

(n! / (n - 5)!) * ((n - 4)! / n!) = 14

The n! terms cancel out:

1 / (n - 5) * (n - 4) / 1 = 14

Simplifying:

(n - 4) / (n - 5) = 14

Now, we can cross multiply:

(n - 4) * 1 = 14 * (n - 5)

Expanding:

n - 4 = 14n - 70

Combine like terms:

0 = 13n - 66

Move the constants to the other side:

13n = 66

Finally, divide by 13 to solve for n:

n = 66 / 13

Simplifying:

n = 5

Therefore, n = 5 is the solution to the equation nP5 = 14(nP4).

To solve for n in the equation nP5 = 14(nP4), where nP5 represents the permutation of n taken 5 at a time, and nP4 represents the permutation of n taken 4 at a time, we can follow these steps:

Step 1: Expand the permutations:
nP5 = n! / (n - 5)!
nP4 = n! / (n - 4)!

Step 2: Substitute these expressions into the equation:
n! / (n - 5)! = 14 * (n! / (n - 4)!)

Step 3: Simplify the equation by canceling out the common terms:
(n - 4)! * n! / (n - 5)! = 14 * n!

Step 4: Cancel out the n! terms on both sides of the equation:
(n - 4)! = 14

Step 5: Solve for (n - 4):
(n - 4)! = 14

To solve this, we need to find the factorial values for numbers greater than or equal to 4 until we find the factorial value that equals 14.

(n - 4)! = 14
(3)! = 14 (not equal to 14)
(4)! = 24 (not equal to 14)
(5)! = 120 (not equal to 14)
(6)! = 720 (not equal to 14)

We stop here since the factorial value of 6 is greater than 14.

Since there is no value of n that satisfies the equation, we can conclude that there is no solution for n in the given equation.