Advanced Higher Maths 2012-2013 : Discussion and Help Thread

But would (n-1)! = (n-1)! x (n-3) also make sense? I understand that is slightly more complicated than the 2010 paper and less likely to be asked.

How could (n-1)! = (n-1)! x (n-3) make sense? (n-1)! appears on both sides!?

How could (n-1)! = (n-1)! x (n-3) make sense? (n-1)! appears on both sides!?

I wanted to say this, but thought it'd be better to let him figure it out.

Oh no, sorry. I didn't mean in the context of that question. I just meant as a general rule does it work? I really am shocking at explaining things, particularly over the internet.

Oh no, sorry. I didn't mean in the context of that question. I just meant as a general rule does it work? I really am shocking at explaining things, particularly over the internet.

No it doesn't work whatever the context. Rules which do work are for example n! = (n-1)! x n, (n-1)! = (n-2)! x (n-1) and (n-2)! = (n-3)! x (n-2) etc. Note the two numbers you are splitting the factorial into must be consecutive.

Why must they be consecutive?

Why must they be consecutive?

Remember with an equation both sides have to be equal to each other? n! is n multiplied by every positive integer from n to 1. This means n! is 1 x 2 x 3 ... x (n - 2) x (n - 1) x n

Look at everything in that sequence after n. It's the same as (n - 1)!. Hence we can get n! = (n - 1)! x n

Applying the same principles to any factorial, say (n-3)!, if you write what that is out, you've got (n - 4)! multiplied by (n - 3) on the end. Hence (n - 3)! = (n - 4)! x (n - 3)

Write the entire sequence out in a line for a couple of factorials, then it'll become obvious why expressing a factorial in the way you've been trying to only works with the consecutive number.

What about this:

(n-1)!=(n-1)!x(n-3)!

What about this:

(n-1)!=(n-1)!x(n-3)!

Why do you think that would work?

Actually nevermind. :P So I just need to know that (n-2)!=(n-2)x(n-3)! works because the 2 and 3 are consecutive numbers?

I'm going to give you a real example. Let's take 6!

This means in n! = 1 x 2 x 3 ... x (n - 2) x (n - 1) x n, n = 6. So we sub all the n's for 6s to get the below.

6! = 1 x 2 x 3 x 4 x 5 x 6

We can insert brackets wherever we want in a product due to the associative law of multiplication.

6! = (1 x 2 x 3 x 4 x 5) x 6

Look at what's inside the bracket, that's the same as 5 factorial, isn't it?

6! = 5! x 6

Now. If we remember that n = 6, if we substitute n in the place for 6, we see:

n! = 5! x n

If n = 6, then (n - 1) = 5, so...

n! = (n - 1)! x n

We can do this for any factorial equation, we've actually proved this type of relationship multiple times algebraically up above. But regardless, let's give you another number example, staying with n = 6

This time we're trying to express (n - 1)! in a different way, and we're letting n = 6 again.

That means (n - 1)! = 5! = 1 x 2 x 3 x 4 x 5

Let's put our brackets in again:

5! = (1 x 2 x 3 x 4) x 5

Look inside the bracket, that's the same as 4!, right?

5! = 4! x 5

As n = 6, (n - 1) = 5, put that back in:

(n - 1)! = 4! x (n - 1)

As (n - 1) = 5, then (n - 2) = 4

(n - 1)! = (n - 2)! x (n - 1)

Do you see why it works with consecutive numbers now that we've used real numbers instead of algebraic notation?

Thanks, I tend to ask questions looking for answers that I sometimes don't actually need for all my subjects, I think it's OCD.

Now express (n - 5)! in a different way.

(n-5)! = (n-5) x (n-6)!

And if n=10 for example that could be written as 5! = 5 x 4!

You've got it!

Sorry to bother you again. I'm having trouble working through 2010 Q5 on the proof. I understand it down to the 3rd line but unsure of the line with the square brackets. I'm just being stupid but I can't figure it out...