Hi,
need to calculate fact(365) but getting #Num!. Any advice?
Thanks

The FACT() function does not work that high.
--
Gary''s Student - gsnu200817


"Jitka" wrote:

Hi,
need to calculate fact(365) but getting #Num!. Any advice?
Thanks

From Excel help "Specifications and limits":
"Largest allowed positive number 1.79769313486231E+308
"
--
David Biddulph


"Gary''s Student" wrote in message
...
The FACT() function does not work that high.
--
Gary''s Student - gsnu200817


"Jitka" wrote:

Hi,
need to calculate fact(365) but getting #Num!. Any advice?
Thanks

Jitka wrote:
Hi,
need to calculate fact(365) but getting #Num!. Any advice?
Thanks

Hi. As a side note, a math program indicates it is 779 digits long.
Here's an approximate value:

2.510412867555873*10^778

As others have mentioned, it's too large for excel.

(If it helps, the Natural Log of 365! is...)

Log(365.!) = 1792.3316495780505
- - -
Dana DeLouis

You can use Excel to get an rough (15 sig fig) approximation to FACT(365) by
trimming down by a couple of orders of magnitude in each row. If B1:B365
contains =ROW(), you can use an array formula =PRODUCT(B1:B365/100) to get
2.51041E+48 or as number with no decimal places to see that the significant
figures are 251041286755588, and then the number would need to be multiplied
by 10^(2*365), so 1E+730 would change the exponent term from E+48 to E+778
(as Dana gave below).
--
David Biddulph


"Dana DeLouis" wrote in message
...
Jitka wrote:
Hi,
need to calculate fact(365) but getting #Num!. Any advice?
Thanks

Hi. As a side note, a math program indicates it is 779 digits long.
Here's an approximate value:

2.510412867555873*10^778

As others have mentioned, it's too large for excel.

(If it helps, the Natural Log of 365! is...)

Log(365.!) = 1792.3316495780505
- - -
Dana DeLouis

Jitka wrote...
need to calculate fact(365) but getting #Num!. Any advice?

Do you really NEED to calculate 365!, or are you working with numbers
of the form n!/[(n-k)!k!] ? If the latter, and if you refuse to use
any of the standard algebraic reductions common in either probability
or combinatorics, you could probably get by using GAMMALN since ln(n!)
= GAMMALN(n+1), so n!/[(n-k)!k!] = ROUND(EXP(GAMMALN(n+1)/GAMMALN(n-k
+1)/GAMMALN(k+1)),0). The ROUND call is needed because Excel's GAMMALN
is only accurate to 10 decimal places and is really poor around 1 and
2 where it should return 0 exactly.

Hi,

Why don't you show us the entire formula you are writing. I doubt you are
just calculating the factorial of 365 (the number of days in the year?)

For example if you are trying to calculate

Pk,n = n!/(n-k)! Then use Excel's PERMUT function

and if you are trying to calcualte

Ck,n = n!/(k!*(n-K)!)

use Excel's COMBIN function.

If this helps, please click the Yes button

Cheers,
Shane Devenshire

"Jitka" wrote:

Hi,
need to calculate fact(365) but getting #Num!. Any advice?
Thanks

Shane Devenshire wrote...
....
For example if you are trying to calculate

Pk,n = n!/(n-k)! * *Then use Excel's PERMUT function

and if you are trying to calcualte

Ck,n = n!/(k!*(n-K)!)

use Excel's COMBIN function.
....

ALWAYS best to test!

Why don't you report back what Excel returns for PERMUT(365,182) and
COMBIN(365,182)?

=COMBIN(365,182) gives me 3.1322E+108
What did your test give, Harlan?
--
David Biddulph

"Harlan Grove" wrote in message
...
ALWAYS best to test!
Why don't you report back what Excel returns for PERMUT(365,182) and
COMBIN(365,182)?
....

"David Biddulph" <groups [at] biddulph.org.uk wrote...
=COMBIN(365,182) gives me 3.1322E+108
What did your test give, Harlan?

Same. What did PERMUT(365,182) give you?

Hi David. That's a great technique. Thanks.

If B1:B365 contains =ROW()...

Perhaps we could eliminate the B1:B365 cells with this Array formula...

=PRODUCT(ROW(INDIRECT("A1:A365"))/100)

Another variation of that same equation is:

=EXP(GAMMALN(366)-365*LN(100))

but it's only good for a few digits, as GammaLn is not very accurate.

Thanks. :)
Dana DeLouis


David Biddulph wrote:
You can use Excel to get an rough (15 sig fig) approximation to FACT(365) by
trimming down by a couple of orders of magnitude in each row. If B1:B365
contains =ROW(), you can use an array formula =PRODUCT(B1:B365/100) to get
2.51041E+48 or as number with no decimal places to see that the significant
figures are 251041286755588, and then the number would need to be multiplied
by 10^(2*365), so 1E+730 would change the exponent term from E+48 to E+778
(as Dana gave below).
--
David Biddulph


"Dana DeLouis" wrote in message
...
Jitka wrote:
Hi,
need to calculate fact(365) but getting #Num!. Any advice?
Thanks
Hi. As a side note, a math program indicates it is 779 digits long.
Here's an approximate value:

2.510412867555873*10^778

As others have mentioned, it's too large for excel.

(If it helps, the Natural Log of 365! is...)

Log(365.!) = 1792.3316495780505
- - -
Dana DeLouis