product of successive integers

[hedgehog]

ok, thanks glenn, and everyone that responded. It's good to find an active forum such as this one.

I'm , perhaps not as skilled as I'd like to be and I am having trouble making "F" double precision. The function gives an "overflow" error for any "n" greater than 12. can anyone give me a hand?

It would still be nice to have the function solve the equation:

x = p! / (n! (p-n)!)

Where the user would be prompted for p and n.

Thanks again,
Hedgehog

[Glenn]

suffixes to "#" and the "F AS LONG" to "F AS DOUBLE." Sorry. I guess I should have realized that even LONG integers wouldn't give you enough range. (I knew there was a reason that when I actually write factorial functions for my own use, I make them double precision. I just didn't remember what it was. And you can probably change the "N AS LONG" to "N AS INTEGER" if you want to--even if you're using my original code with LONG integers. The only exception would be if you want the factorial of a number larger than 32,767--and I don't see that happening.)


Well, unless I made some more typos doing things on the fly again, this should do it.



DECLARE FUNCTION FACT#(N AS INTEGER)
DIM N AS INTEGER
INPUT "N";N
PRINT LTRIM$(STR$(N));" = ";FACT(N)
END
'
' The above was just a "driver." Here's the function.
'
FUNCTION FACT#(N AS INTEGER)
DIM F AS DOUBLE,I AS INTEGER
F=1#
IF N>0 THEN
FOR I=1 TO N
F=F*I
NEXT I
ELSEIF N<0 THEN
PRINT "CAN'T DO NEGATIVE NUMBERS. LOOK UP THE 'GAMMA FUNCTION.'"
STOP
END IF
FACT=F
END FUNCTION