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Ops sorry the calling should be (permutate '(1 2 3) 2), with the 2nd parameter being a number, where it will determine the number of permutates.

Thanks for all the help. I have worked out a solution on my own, and it is some what similar. Here's my solution to share: (define (permutate lst len) (cond ((= 0 len) '(())) (else (append-map (lambda (x) (map (lambda (y) (reverse (cons y x ))) lst)) (permutate lst (- len 1)))))) And similarly i do ...

Hi thanks for the reply. But actually i can't use the function in mention. Hmmm, what i would like to do is to produce a permutation. Let's take alphabets for a-z for example. When I call my function (permutate n): (permutate 1) output: (("a")("b")("c")....("x"...

Hi mates, I'm trying to produce and algorithm that counts (base 10) and display in such a manner: Numbers 0 to 9 will produce this output: ((0)(1)(2)(3).....(8)(9)) Numbers 10 to 20 will produce this ouput: ((00)(01)(02)...(09)(10)(11)...(20)) The thing is each time the number gets larger, I would l...