# Discrete Math Question

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1. Nov 12, 2014

### hlzombi

Prove the following theorem:
Theorem For a prime number p and integer i,
if 0 < i < p then p!/[(p− i)! * i] * 1/p

Not sure how to go about this. I wanted to do a direct proof and this is what Ive got so far.
let i = p-n
then p!/[(p-n)!*(p-n)] but that doesnt exactly prove much.

2. Nov 12, 2014

### RUber

hlzombi,
First, please note that you should use the template and this seems like it might be better to place in the math forum.
$\frac{p!}{(p-n)!*(p-n)}=\frac{p*(p-1)*...*(p-n+1)}{p-n}$
Finally, I am not sure what you are asked to prove...is there some equality or property here?
You need to be more clear.

3. Nov 12, 2014

### hlzombi

Apologies, I'm new here. I tried to follow the template as best as I could.

To clarify, I'm trying to prove the theorem p!/[(p− i)! * i] * 1/p where 0<i<p when p is a prime number and i is an integer.

4. Nov 12, 2014

### pasmith

Your statement is incomplete. What are you trying to prove about $\frac{p!}{(p-i)!i} \times \frac 1p$ when $p$ is prime and $i$ is an integer?

5. Nov 12, 2014

### hlzombi

Im trying to verify the theorem under those conditions

6. Nov 12, 2014

### RUber

You did not write the theorem.

7. Nov 12, 2014

### hlzombi

8. Nov 12, 2014

### RUber

That says p divides p choose i. That is not what you wrote above.

9. Nov 12, 2014

### RUber

To demonstrate this, you can use induction.
Show that for a base case (i=1) $p\left| \left( \begin{array}{c} p\\i\end{array}\right) \right.$
Assume for some n < p-1, the statement holds.
Show that $p\left| \left( \begin{array}{c} p\\n+1 \end{array}\right) \right.$

10. Nov 12, 2014

### Staff: Mentor

When you use the template, don't delete its three parts.

Also, I moved this thread, as it was better suited in one of the math sections.