Hi,
The answer actually depends on whether the order is important (ie if it
matters if the user selects from x from the first menu and y from the
second, and if this is deemed to be identical to selecting y from the first
and x from the second).
If the order is important, then the number of permutations is
(dp_items)!/(dp_items-dp_menu)! where ! is the factorial symbol eg
6!=6*5*4*3*2*1
In the example below, this gives 5!/1!=120
If the order is unimportant, the the number of combinations is
(dp_items)!/((dp_menu)!(dp_items-dp_menu)!)
In the example below, this gives 5!/4!1! = 5
For the lottery below, the answer is actually 56!/6!50! = 32468436:1 (which
is pretty darn close to zero!)
:-)
Nigel Russell
Linnsell Science Software
-----Original Message-----
From: Paul A. Procacci [mailto:[EMAIL PROTECTED]]
Sent: Thursday, July 19, 2001 5:44 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Re: OT Math Question
Well, If I'm right. The total number of posible combinations in the NJ
state lottery, assuming there are, say, 56 numbers is
var total = 56;
total*(total-1)*(total-2)*(total-3)*(total-4)*(total-5) == 23377273920:1
Now that assuming one number was pulled from the pot and the same number
didn't exists in the pot. Now, please anyone correct me if I'm wrong,
but I believe the solution to rm's quesion is as follows:
var dp_menus = 4 // Number of drop-down menus
var dp_menus = 5 // Number of dp_items per menu
dp_menus^dp_items == 1024:1 //Total number of possibilities
Does that look right? I hope this helps
Paul.
Rm wrote:
>
> if i have four drop down menus with five values in
> each, and the values are the same for each of the four
> drop down menus, how many variations are there without
> duplicates, any dups,
>
> I though the formula was 5 to the 4th power minus 5,
> this can't possibly be right.
>
> math impaired and trying to fake it....
>
> rm
>
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