1. ## [SOLVED] Permutations question

nP3 = 60
????
so
n!/(n-3)!

=

n(n-1)(n-2)(n-3)....1
over
(n-3)(n-4)(n-5)....1

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

=
n^3-3n^2+2n-60=0

So what do you do polynomial division?
what if its like nP20
is there something Im doing wrong?
thanks.

2. Originally Posted by brentwoodbc
nP3 = 60
????
so
n!/(n-3)!

=

n(n-1)(n-2)(n-3)....1
over
(n-3)(n-4)(n-5)....1

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

=
n^3-3n^2+2n-60=0

So what do you do polynomial division?
what if its like nP20
is there something Im doing wrong?
thanks.
Yep you do polynomial division, I'm not sure how you were taught. I was taught to try and find a factor.

f(5) = 125-75+10-60 = 0.

Therefore (n-5) is a factor and we can divide through using long division and solve the remaining quadratic.

or compare coefficients: (n-5)(an^2+bn+c) = n^3-3n^2+2n-60

For example compare coefficients of x^0 (the constant): -5c = -60 therefore c=12

3. Originally Posted by e^(i*pi)
Yep you do polynomial division, I'm not sure how you were taught. I was taught to try and find a factor.

f(5) = 125-75+10-60 = 0.

Therefore (n-5) is a factor and we can divide through using long division and solve the remaining quadratic.

or compare coefficients: (n-5)(an^2+bn+c) = n^3-3n^2+2n-60

For example compare coefficients of x^0 (the constant): -5c = -60 therefore c=12
ya same here find a factor.
darn, I hate that. thanks.

4. also I dont know what I'd do if I got an n^4 term because I was only taught how to do cubic functions, I was told you cant don quardics with polynomial division because of finding the factors etc.

5. Originally Posted by brentwoodbc
also I dont know what I'd do if I got an n^4 term because I was only taught how to do cubic functions, I was told you cant don quardics with polynomial division because of finding the factors etc.
You can do much the same thing for a quartic equation you just have to hope you get lucky with a rational solution.

There is also a special case which is easy to solve - these are of the form $ax^4+bx^2+c=0$ which can be solved like a quadratic

6. thanks I dont think we will get anything more than an x^3 anyways, probly just alot of nP2's

7. Ive done a couple using long division, and the remainder = 0 so that is good....But the quadratics I get are not factorable nor can you use the quadratic equation because the discriminant is negative?

for the example I gave I got n^2+2n+12?
Edit.
I guess it means the answer is the one factor of 5.

8. Originally Posted by brentwoodbc
Ive done a couple using long division, and the remainder = 0 so that is good....But the quadratics I get are not factorable nor can you use the quadratic equation because the discriminant is negative?

for the example I gave I got n^2+2n+12?
Edit.
I guess it means the answer is the one factor of 5.
It means the other two solutions are complex numbers, if you've not learned about them yet then put something like "n=5 is the only real solution"

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# (n-1)(n-2)(n-3)! factorizar

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