Hi Team,

Such problems like yours are usually being solved with recursion
algorithms. It is very easy with recursion. If you manage to solve it
with recursion in JBase please send me the algorithm written.

I’ll try to picture you the iterative algorithm.
For 4 symbols you have to have 24 combinations. Right.
OK, here is the logic
We start with your 6 permutations without repetitions which are
actually combinations.

 ABC
 ACB
 BCA
 BAC
 CAB
 CBA

Let’s add one more symbol D.
If you add D at the first position you have the following sequence
DABC
DACB
DBCA
DBAC
DCAB
DCBA

If you continue to add D on the next positions, i.e. 2th ,3th and 4th
you are going to produce the following combinations

ADBC
ADCB
BDCA
BDAC
CDAB
CDBA

ABDC
ACDB
BCDA
BADC
CADB
CBDA

ABCD
ACBD
BCAD
BACD
CABD
CBAD

So you now have 24 combinations. With 5 symbols you’ll have 120
combinations. I think it is totally clear now.
Can you please send me the algorithm when you’ll manage to develop it?


Kind regards
Igor Micev

On May 8, 8:06 am, kavi <[email protected]> wrote:
> Hi Team,
>
> Could anyone please give me the logic for the following
>
> 1. I have an array of values with VM seperation.The array can have any
> number of values.
>
> Example ---- ARRAY = 'A':VM:'B':VM:'C'
>
> 2. If the count is 3, i want 3!(factorial) combinations to be
> displayed like
>
> ABC
> ACB
> BCA
> BAC
> CAB
> CBA
>
> Likewise, if the count is 4, then there should be 4! combinations (ie
> 24 combinations).
>
> The count should be n numbers.(Generic and not a specific value.)
>
> I need a logic to implement this. Please help me

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