Quick question on Factorials

[vorcil]

I need to figure out the following factorial

\frac{297!}{98! * 199!}

then take the logarithim of that

Is there a rule that I can use to simplify the equation and get the same result?

,

I did another example where I used

\frac{310!}{2!*299!}
and I figured it out to be
(310*309*308*307*306*305*304*303*302*301*300) / 2!

but if i were to apply the same rule
I'd need to do 98 multiplications starting from 297 going down to 199
and that'd take way too long in my calculator. i.e 297*296*295....200 / 98!

please help, I need some rules to follow
i couldn't find any anywhere,

[Dickfore]

Stirling's formula:


n! \approx \sqrt{2 \, \pi n} \left(\frac{n}{e}\right)^{n}, \ n >> 1

[Hurkyl]

Could you be more precise on what you mean by "figure out"?

[vorcil]

Could you be more precise on what you mean by "figure out"?



well I used a big number calculator that lets me use factorials up to 500!,

found out the answer to be 45.4 or something

then tried to replicate that answer on my normal calculator by guessing that

310!/299! is actually 300!*301!*...310!, then divided that by 2! which is equal to 2,

and it was the same answer, 45.4


thanks to that other guy,
I forgot about stirlings approximations :P

[Mark44]

well I used a big number calculator that lets me use factorials up to 500!,

found out the answer to be 45.4 or something

then tried to replicate that answer on my normal calculator by guessing that

310!/299! is actually 300!*301!*...310!, then divided that by 2! which is equal to 2,

and it was the same answer, 45.4
Probably a typo, but 310!/299! = [310*309*308*307*306*305*304*303*302*301*300*299!]/299!.

The 299! factors cancel and you're left with 310*309*308*307*306*305*304*303*302*301*300.



thanks to that other guy,
I forgot about stirlings approximations :P