This is actually part of a statistics problem I'm working but I forgot how to do this and if someone can refresh my memory, I might be able to solve these.

The problem is with factorials.

20! / 5! (20-5)!

I know I can reduce and cancel, but how and where?

If I do
20! / 5! (15)! I get 20! / 75!
Here's where I think I'm screwing up. I think I can cancel the 15! on the bottom and 15! from the top, but then I'm left with 20*19*18*17*16=1860480.
Calculate out the 5! on the bottom, I get 120. Divide the two and I get 15504. Is this correct? Is there an easier way?

20!/[5!*15!) = 20*19*18*17*16/(1*2*3*4*5)

= 20*19*18*17*16/120
= 19*18*17*16/6
= 19*3*17*16 = 15504

Your answer is correct. The only easier way is to use a calculator.

the mass of a neutron is 1.67(10^(-27))kg. what is the mass of 7.03(10^(21))neutrons?

To simplify the expression 20! / (5! * (20-5)!), you can use the properties of factorials.

First, let's simplify each factorial individually:

- 20! means multiplying all the numbers from 20 down to 1: 20 * 19 * 18 * ... * 3 * 2 * 1.
- 5! means multiplying all the numbers from 5 down to 1: 5 * 4 * 3 * 2 * 1.
- (20-5)! means multiplying all the numbers from 15 down to 1: 15 * 14 * 13 * ... * 3 * 2 * 1.

Now, let's cancel out common terms.

The 20! in the numerator can be written as (20 * 19 * 18 * ... * 3 * 2 * 1).

The 5! and (20-5)! in the denominator can be multiplied together to give (5 * 4 * 3 * 2 * 1) * (15 * 14 * 13 * ... * 3 * 2 * 1).

Notice that all terms from 1 to 5 cancel out in the numerator and denominator, so we are left with:

(20 * 19 * 18 * 17 * 16 * 15 * 14 * 13 * ... * 3 * 2 * 1) / (15 * 14 * 13 * ... * 3 * 2 * 1).

After canceling out these common terms, we are left with:

20 * 19 * 18 * 17 * 16 = 1860480.

So, the answer is indeed 1860480.

This method of canceling out common terms saves you from having to calculate the entire factorials, which can be time-consuming for larger numbers.