I knew there was a reason I shouldn't have been sleeping in Math class. Damn.
I figured from the repeating patterns in the result that there was most likely a mathematical solution.
The fact that you know this explicitly really humbles me.
I am going to study this and try and learn something from it.
Thanks,
Tom
On Nov 22, 2003, at 11:04 AM, Geoff Canyon wrote:
(Anyone not interested in math hit delete right now)
I should have pointed out the formula for this. When choosing y objects from x possible objects, the number of combinations xCy is
(x!)/(y!)*((x-y)!)
Where x! means x * (x-1) * (x-2) * ... * 2 * 1
In this case that translates to
7!/3!*(7-3)!
7!/3!*4!
The nice thing about this formula is that whichever of the divisors is larger simple cancels out that portion of the dividend. So:
7!/3!*4!
which is
7*6*5*4*3*2*1/3*2*1*4*3*2*1
becomes simply
7*6*5/3*2*1
3*2 cancels the 6, and you have 7*5 or 35
If you _do_ care about what order the numbers come in, that's permutations, and the formula is similar but simpler:
xPr = x!/(x-y)!
So if you wanted all the ways you could arrange 3 numbers from 7 that would be
7!/(7-3)!
or
7!/4!
which is 7*6*5, or 210
regards,
Geoff Canyon [EMAIL PROTECTED]
On Nov 22, 2003, at 12:19 AM, Geoff Canyon wrote:
To verify that there are 35 solutions, consider that this problem translates to: choose three numbers from the set "1,2,3,4,5,6,7" To solve that, find 7C3, which is (7*6*5)/(3*2*1) That's 210/6, or 35
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