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Quick question on Factorials

  1. May 26, 2010 #1
    I need to figure out the following factorial

    [tex] \frac{297!}{98! * 199!} [/tex]

    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

    [tex] \frac{310!}{2!*299!} [/tex]
    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,
     
  2. jcsd
  3. May 26, 2010 #2
    Stirling's formula:

    [tex]
    n! \approx \sqrt{2 \, \pi n} \left(\frac{n}{e}\right)^{n}, \ n >> 1
    [/tex]
     
  4. May 26, 2010 #3

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Could you be more precise on what you mean by "figure out"?
     
  5. May 27, 2010 #4


    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
     
  6. May 31, 2010 #5

    Mark44

    Staff: Mentor

    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.
     
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