1. Stirling and factorials

Okay so I have 1000 houses in an area where lightning strikes one house at random every week, and I need to approximate the probablity of lightning striking the same house twice (or more, presumably) within one year (52 weeks), and then for two years. I'm allowed to use the Stirling Formula: $\displaystyle \sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}<n!<\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n+\frac{1}{12(n-1)}}$.

Now if I did this right, with n multichoose k = $\displaystyle \left(\begin{array}{c} n+k-1\\ k\end{array}\right)$ I have $\displaystyle \left(\begin{array}{c} 1051\\ 52\end{array}\right)$ possible outcomes with $\displaystyle \left(\begin{array}{c} 1051\\ 52\end{array}\right)-\left(\begin{array}{c} 1000\\ 52\end{array}\right)$ favorible outcomes which expanding with $\displaystyle \left(\begin{array}{c} n\\ k\end{array}\right)=\frac{n!}{(n-k)!k!}$ and then dividing leaves me with $\displaystyle \frac{1000!999!}{948!1051!}$(kind of messy). Is there some way to further simplify this before plugging it into Stirling, or perhaps a simpler way of going about this altogether?

2. Not as messy as you think...

$\displaystyle \frac{1000!999!}{948!1051!}$ $\displaystyle = \frac{999*998*997*...*950*949}{1051*1050*...*1002* 1001}$ $\displaystyle =\frac{949}{1001}*\frac{950}{1002}*...*\frac{999}{ 1051}$ $\displaystyle = \prod_{n=1}^{51}\frac{948+n}{1000+n}$ $\displaystyle \approx 7.04\%$

This can be done with a simple program on a graphing calculator. (I did not check your work up to this point, but 7% seems a bit high for the likelihood of getting struck by lightning twice - or maybe I've just been lucky.)

3. The idea is to do it without a computer. With a normal calculator at most.

4. Originally Posted by realpart1/2
The idea is to do it without a computer. With a normal calculator at most.
With apologies to Keith Devlin, I say that is like the local vocational education department insisting that horse shoeing is an in demand job.

5. Originally Posted by Plato
With apologies to Keith Devlin, I say that is like the local vocational education department insisting that horse shoeing is an in demand job.
I don't make the rules, I'm just bound to them.

6. Originally Posted by realpart1/2
I don't make the rules, I'm just bound to them.
Only if you choose to be.
Students can walk away from a outdated program.

7. Originally Posted by Plato
Only if you choose to be.
My professor's ability to fail me is a pretty good motivation. Now, can I get some genuine help on this?

8. The Fundamental Counting Principle

If I'm reading this problem right, you shouldn't be using the formula for $\displaystyle _nC_r$ at all.

Picture yourself living in one of the 1000 houses. Every week, you have a .1% chance of getting hit, and a 99.9% chance of not getting hit. So your chances of not getting hit two weeks in a row is $\displaystyle (.999)^2\approx$99.8%. So after 52 weeks, you have a $\displaystyle (.999)^{52} \approx$ 94.93% chance of not getting hit, and a 5.07% chance of getting hit. Wouldn't then the chance of getting hit twice in that same timeframe be $\displaystyle (.0507)^2 \approx$ .26% ?