JenniferMurphy
Well-known Member
- Joined
- Jul 23, 2011
- Messages
- 2,687
- Office Version
- 365
- Platform
- Windows
As I understand it, the combin & permut functions only calculate the results without replacement. There are no corresponding functions for with replacement. Is that correct?
The formula for permutations with replacement is just n^k, but the one for combinations is ugly: (n+k-1)!/k!(n-1)!.
Do I need to write a UDF to avoid having to enter that each time and probably make a typo 30% of the time?
Would it have killed the M$FT developers to have added a third parameter to these functions, something like 0/1 where 0 = without replacement and 1 = with? They could have set the default to without replacement so that there would be no compatibility problems. Seems obvious and simple.
The formula for permutations with replacement is just n^k, but the one for combinations is ugly: (n+k-1)!/k!(n-1)!.
Do I need to write a UDF to avoid having to enter that each time and probably make a typo 30% of the time?
Would it have killed the M$FT developers to have added a third parameter to these functions, something like 0/1 where 0 = without replacement and 1 = with? They could have set the default to without replacement so that there would be no compatibility problems. Seems obvious and simple.