Question 1164031
<pre>
Ikleyn hasn't noticed that 940! and 910! have over 2000 digits each and that
is beyond the capacity of all calculators.  

a) {{{940!/(30! 910!)}}}

{{{X}}}{{{""=""}}}{{{940!/(30!910!)}}}

Take logs (base ten)

{{{log(X)}}}{{{""=""}}}{{{log((940!/(30!910!)))}}}

{{{log(X)}}}{{{""=""}}}{{{log(940!)-log(30!)-log(910!)}}}

We use Gosper's improvement of Stirling's approximate formula for n!:

n! = √[(2n+1/3)∙π]∙n<sup>n</sup>∙e<sup>-n</sup>

We take the log<sub>10</sub> of that:

log(n!) = 0.5∙log[(2n+1/3)∙π]+n∙log(n)+(-n)/log(e)

We find by calculator using the above formula:

{{{log(940!)}}}{{{""=""}}}{{{2388.3890617535}}} 

{{{log(910!)}}}{{{""=""}}}{{{2299.3983388775}}}

{{{log(30!)}}}{{{""=""}}}{{{32.42366007}}}

[We didn't need Gosper's formula for that last one, for it isn't too large
for the calculator.]

Substituting in

{{{log(X)}}}{{{""=""}}}{{{log(940!)-log(30!)-log(910!)}}}

{{{log(X)}}}{{{""=""}}}{{{56.56706293}}}

{{{X}}}{{{""=""}}}{{{10^56.56706293}}}

{{{X}}}{{{""=""}}}{{{10^56*10^0.56706293}}}

{{{X}}}{{{""=""}}}{{{10^56*3.690310679}}}

Answer:  3.6903107 × 10<sup>56</sup>

Do the others the same way.  Use logs base ten.  Use Gosper's approximation
formula or whatever approximation formula your teacher told you to use.  

Edwin</pre>