find the value of n in the equation nP3=210?

Question

pleaso show your solutions, solutions are a must.

Answer 1

Remember that the formula for nPr is

nPr = n! / (n - r)!

That means

nP3 = n! / (n - 3)!

And, since nP3 = 210,

210 = n! / (n - 3)!

Note that n! can be defined recursively;
n! = n(n - 1)!
n! = n(n - 1)(n - 2)!
n! = n(n - 1)(n - 2)(n - 3)!

Applying this to n!, we get

210 = n(n - 1)(n - 2)(n - 3)! / (n - 3)!

Cancelling the (n - 3)! on the top with the (n - 3)! on the bottom, we get

210 = n(n - 1)(n - 2)

Now, we expand the right hand side.

210 = (n^2 - n)(n - 2)
210 = n^3 - 2n^2 - n^2 - 2n
210 = n^3 - 3n^2 - 2n

Moving the 210 over to the right hand side,

0 = n^3 - 3n^2 - 2n - 210

Now, if we let p(n) = n^3 - 3n^2 - 2n - 210, we determine how to solve this cubic by plugging in factors of 210 for p(n).

Once you find a root r that works, (n - r) will be a factor, and you use long division to factor accordingly.

Answer 2

n^3=210

n=210^(1/3)