This does not work (365!/335!)/365^30 I get internal error

In fact, there is a way to get the calculator here to calculate this (using the npr function):

$${\frac{{\left({\frac{{\mathtt{365}}{!}}{({\mathtt{365}}{\mathtt{\,-\,}}\left({\mathtt{365}}{\mathtt{\,-\,}}{\mathtt{335}}\right)){!}}}\right)}}{{{\mathtt{365}}}^{{\mathtt{30}}}}} = {\mathtt{0.293\: \!683\: \!757\: \!280\: \!731\: \!3}}$$  

I entered the above as npr(365,(365-335))/365^30.

Yep, way to large!

Basically, infinity/infinity^(365^30) which is extremely large.

((((infinity)/(infinity)))/73924080909700308571344669689235259082192300936032301233150064945220947265625)

Notice that (365!/335!) =

(365*364*363*362..........338*337*336)

Note that there are 30 terms in the above.

And each one of these individual terms would be divided by 365.

So we would have

(365/365)*(364/365)*(363/365)*(362/365)........(338/365)*(337*365)*(336/365)

And note that the first term is 1, but all the rest of the terms are <1.

And the result of the last division (336/365) = .9205479452054795

And since this is being multiplied by all the preceding multiplications whose total product is  < 1, we can say that the result of this expression is < 0.9205479452054795

(In fact, it's far less....)

I leave it to you to find the exact result......!!!!

Interesting explanation, Phill! So small, the calculator couldn't catch it!

The result isn't as small as you might expect:

Thanks for that answer, Alan........very imformative !!!

It looks like you "tricked" the calculator by using the permutation function!!!

Nice work......