Technically, he's asking for a combination, which would be
(x choose y)
which is
x!/y!(x-y)! (number of combinations)
instead of
x!/(x-y)! (number of permutations, in which order counts)

So if there are 7 numbers in the original list, and you need 6 of them, 
it's
7!/6!(7-6)! = 7!/6! * 1! = 7!/6! = 7
for 7 numbers, choose 5
7 choose 5 = 7!/5!(7-5)! = 7!/5!*2! = 7 * 6 / 2 = 7 * 3 = 21

Hopefully, that made some sense.

I missed that part in my original post.

--Ben Doom

Christopher Jordan wrote:
> I'm curious why in your case 1,2,3,4,5,6 is considered to be the same as
> 1,2,3,4,6,5?  I was going to suggest that you use n! to figure out how many
> possible permutations of the string there were, but that wouldn't work given
> your requirements.
> 
> Chris
> 
> 
> On 9/25/07, Ken Fassman <[EMAIL PROTECTED]> wrote:
>> Hi,
>> I have a problem I'm struggling to solve in ColdFusion - I wonder if
>> someone
>> can help me with.
>>
>> I am being given a variable length, numeric string (min 6, max 20).  I
>> need
>> to show all combinations of those values that will result in a unique set
>> of
>> 6.  The resulting output is sorted - so 1,2,3,4,5,6 is the same as
>> 1,2,3,4,6,5.
>>
>> I'd also like to see if there is an easy way to calculate the expected
>> number of each in advance.
>>
>> Just to illustrate what I'm looking to do (in case its not clear from
>> above):
>>
>> The input string might be "1,2,3,4,5,6,7" and I'd need to return the
>> following results:
>> 123456
>> 123457
>> 123467
>> 123567
>> 124567
>> 134567
>> 234567
>>
>>
>> Any thoughts on this on how to approach this are greatly appreciated.
>>
>> Thanks!
>>
>> - KsF
>>
>>
>>
> 
> 

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