Why is nPr represented as n! / (n-r)!?

Question

I recently started re-learning maths and have started with Combinatorics. I was trying to understand permutations and noticed that nPr which is number of ways r elements can be picked and arranged out of n elements is equal to n! / (n-r)!.

When I was going through the proof, I realised it's because of the multiplication n * (n-1) * ... * (n-r+1)

My questions is - since factorials can not be calculated without going through each multiplication, what are we gaining by denoting value of nPr in the 1st form instead of the 2nd.

• First - n! / (n-r)! - Communicates that an extra work of calculating (n-r)! is needed.
• Second - n * (n-1) * ... * (n-r+1) - No extra work and easy to understand.

Am I missing something?

Answer 1

Depends on what you mean by "extra work". The variant with factorials is certainly easier to write by hand. And on a calculator with factorials it is much easier to type, significantly reducing the risk of human error.

Answer 2

No I do not think you are. But in practice, - your calculation is what we all, - and our calculators does.

The notation is just a lot easier. As you have demonstrated. :)

Answer 3

In maths, different presentations of the same thing might yield different insights. In this case, you might derive $^nP_r$ from $^nC_r$ as follows:

We know that $^nC_r$ counts the number of ways to choose $r$ objects from $n$ objects, without caring about the order you choose them from.

But suppose now we want to care about the order. There are $r!$ ways to order the $r$ chosen objects. So we have $$^nP_r = {}^nC_r\times r!$$

Recall the formula for

$$^nC_r=\frac{n!}{r!(n-r)!}$$

so we have $$^nP_r=\frac{n!}{(n-r)!}$$