Separating (a+b)! into a! and b!?

Question

Is it possible to separate (a+b)! into terms strictly involving a! and b!?

(Obviously (a+b)! = a!b![(a+b)!/a!b!]. I'm looking for something like (a+b)! = a!b!*k, where k is also a function of strictly a! and b!)

Answer 1

(a+b)C(a) = (a+b)! / [a! b!]
So (a+b)! = a! b! * (a+b)C(a)

That's the only way

Answer 2

Ok, to start, let's think about the most natural way to expand (a+b)!. It would be nice if we could separate it into (a+b)! = a! + b! but clearly this doesn't work. So let's try a! * b!

If a = 3, b = 2, then (a+b)! = 5! = 5 * 4 * 3 * 2 *1
Now a! * b! = (3 * 2 * 1) * (2 * 1)

To correct for this, we will need to get rid of the extra 2. That's easy, simply multiply by b/b!, or some variation of that. But how to correct for the 5? You will quickly see that it is simply not possible to do, because the "gap" between them increases at a factorial rate and we need to some form of (a x b)! (where x is an operator) to correct for this.

There are, however, some interesting ways given a (a x b)! form, though I don't think they are what you are looking for.

For example, (a - b)! b! would perform the operation without using (a + b)! (this works because ! is undefined for negative numbers and 'a' will be greater than or equal to 'b').

Answer 3

I don't believe so, but cannot prove it. Interesting problem.

Answer 4

i think not possible

Answer 5

No I suppose not.