Simplyfing Probability equation

Question

I was solving a homework problem, and I had obtained a formula for the required probability in the question.
What I wanted to ask could it be more simplified? $$P = \sum_{i=0}^{a}( \frac{a!}{(a-i)!} * \frac{(s-i-1)!}{s!})$$

Answer 1

The answer is in fact $P=1/(s-a)$. Here's why.

Rewrite the sum as

$$\begin{align} P &= \frac{a!}{s!} \sum_{i=0}^a \frac{(s-i-1)!}{(a-i)!} \\ &= \frac{a!}{s!} \sum_{i=0}^a \frac{i}{s-i} \frac{(s-i)!}{(a-i)!} \\ &= \frac{a!}{s!} \sum_{i=0}^a \frac{1}{s-a+i} \frac{(s-a+i)!}{i!} \\ &= \frac{1}{\binom{s}{a}} \sum_{i=0}^a \frac{1}{s-a+i} \binom{s-a+i}{i} \\ &= \frac{1}{\binom{s}{a}} \frac{1}{s-a} \underbrace{\sum_{i=0}^a \binom{s-a-1+i}{i}}_{\text{This is a well-known sum}}\\ &= \frac{1}{\binom{s}{a}} \frac{1}{s-a} \binom{s}{a} \\ &=\frac{1}{s-a} \end{align}$$

The well-known sum may be found in, for example, Concrete Mathematics by Knuth, Graham, and Patashnik, Second Ed., p. 161.

Answer 2

It could be more simplified in this way: $$P = \frac{a!}{s!} \sum_{i=0}^{a}\frac{(s-i-1)!}{(a-i)!}$$ not more