Question about factorials...

[Random Variable]

For the second proof of Proposition 7.2.11, I don't understand the step in which did they reduced the two fractions into one fraction.

http://ndp.jct.ac.il/tutorials/Discrete/node82.html

 

[ActuaryTm]

Reduce the denominator of each set of terms on the right side of the equation.

p!(n-p)(n-p-1)! is equivalent to p!(n-p)!.

p(p-1)!(n-p)! reduces to p!(n-p)! as well.

Now both terms have a common denominator; thus, one can add the numerators. Factor out an (n-1)! from the numerator to acheive the final numerator listed in the fourth step.

 

[Random Variable]

So (n-p)(n-p-1)! = (n-p)! and p(p-1)! = p! ?

EDIT: That does actually make sense.

 

[ActuaryTm]

Originally posted by: Vespasian
So (n-p)(n-p-1)! = (n-p)! and p(p-1)! = p! ?

Precisely - at least within the contraints of the given of the proposition: 0<=p<=n. Regard the definition of a factorial, and you should be able to deduce why. Or, merely look at an example:

(9-4)(9-4-1)! = (9-4)!
5(4)! = 5!
5(4*3*2*1) = 5!
5! = 5!