set - Understanding A bit of Haskell -

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i have quick question haskell. i've been following learn haskell, , bit confused execution order / logic of following snippet, used calculate side lengths of triangle, when sides equal or less 10 , total perimeter of triangle 24:

[(a,b,c) | c <- [1..10], b <- [1..c], <- [1..b], a^2 + b^2 == c^2, a+b+c==24]

the part confusing me upper expansion bound on b , a binding. gather, ..c , ..b used remove additional permutations (combinations?) of same set of triangle sides.

when run ..c/b, answer:

[(6,8,10)]

when don't have ..c/b:

[(a,b,c) | c <- [1..10], b <- [1..10], <- [1..10], a^2 + b^2 == c^2, a+b+c==24]

as didn't when typed in, got:

[(8,6,10),(6,8,10)]

which representative of same triangle, save a , b values have been swapped.

so, can walk me through logic / execution / evaluation of what's going on here?

the original version considers triplets (a,b,c) c number between 1 , 10, b number between 1 , c , number between 1 , b. (6,8,10) fits criteria, (8,6,10) doesn't (because here 8 , b 6, isn't between 0 , 6).

in version consider triplets (a,b,c) a, b , c between 1 , 10. make no restrictions on how a, b , c relate each other, (8, 6, 10) not excluded since numbers in indeed between 1 , 10.

if think of in terms of imperative for-loops, version this:

for c 1 10:   b 1 10:     1 10:       if a^2 + b^2 == c^2 , a+b+c==24:         add (a,b,c) result

while original version this:

for c 1 10:   b 1 c:     c 1 b:       if a^2 + b^2 == c^2 , a+b+c==24:         add (a,b,c) result

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