Mirko Rahn <[EMAIL PROTECTED]> writes:
(snip)
> Correct (and more natural):
>
> nOf 0 _      = [[]]
> nOf n (x:xs) = map (x:) (nOf (n-1) xs) ++ nOf n xs
> nOf _ []     = []

Thanks very much - in both claims you're indeed correct.

-- Mark

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