# Factorial Algebra

1. Dec 25, 2008

### DecayProduct

1. The problem statement, all variables and given/known data

P(n, 4) = 40[P(n-1, 2)]

2. Relevant equations

???

3. The attempt at a solution

I boiled this down to the equation n!/(n-4)! = 40[(n-1)!/(n-3)!]. The problem is, I have no idea how to perform the correct operations on these factorials. I found the answer to be n = 8, but this was just trial and error. The factorials don't seem to behave like regular numbers, where cross multiplication or LCD would work. The text book this comes from is old, and only glances over the factorials. I suppose a teacher's edition would provide more info. Any hints or tips, as usual, are appreciated.

2. Dec 25, 2008

### Avodyne

Just use the definition of the factorial to simplify:

$$n! = n(n{-}1)(n{-}2)\ldots \cdot 2\cdot 1$$

$$(n{-}4)! = (n{-}4)(n{-}3)\ldots \cdot 2\cdot 1$$

So

$$n!/(n{-}4)! = n(n{-}1)(n{-}2)(n{-}3).$$

Similarly on the right-hand side. Now cancel common factors. You should end up with a quadratic equation in n, with only one positive integer solution.

3. Dec 26, 2008

### DecayProduct

I'm sorry, I guess I'm just thick! My first experience with factorials is with this problem, so I just don't get it. Since I don't know n, how can I know how far to carry the (n-x)? As in here when you you say: $$n! = n(n{-}1)(n{-}2)\ldots \cdot 2\cdot 1$$

I mean, n could be anything, so (n-1), (n-2),... (n-500)? An explanation of the factorial definition you stated would be great.

4. Dec 28, 2008

### yeongil

Okay then, here's n! and (n-4)! again, but I expanded Avodyne's definition:
$$n! = n(n{-}1)(n{-}2)(n{-}3)(n{-}4)(n{-}5)\ldots \cdot 2\cdot 1$$
and
$$(n{-}4)! = (n{-}4)(n{-}5)\ldots \cdot 2\cdot 1$$

Now,
$$\frac{n!}{(n{-}4)!} = \frac{n(n{-}1)(n{-}2)(n{-}3)(n{-}4)(n{-}5)\ldots \cdot 2\cdot 1}{(n{-}4)(n{-}5)\ldots \cdot 2\cdot 1}$$

Notice how the factors (n - 4), (n - 5), all the way to 1, cancel out? So you're left over with

$$\frac{n!}{(n{-}4)!} = n(n{-}1)(n{-}2)(n{-}3).$$

Now try with the right hand side and see what you get.

I actually ended up with a quartic, not a quadratic, with three positive roots and one negative one. The negative solution can be discarded, and so can two of the positive solutions (because they are small enough that makes (n-4) negative), which means the remaining positive number is the answer (which, as you said, is n=8).

01

Last edited: Dec 28, 2008