using DeleteCases with CoprimeQ

Question

first let me show what I have working correctly

f = Permutations[Range[5], {3}]
Riffle[f, Apply[CoprimeQ, f, {1}]]

now I would like to automate the deletion of a list {a,b,c} that returns false from this test. I can probably do it with Do[] but I wanted to learn functional programming the right way using built in functions and pure function on one line of code.

I had spent a lot of time trying to follow this example for DeleteCases[] here but it seems that it requires building a custom function. it also does not perfectly correlate to my example because in that tutorial they are comparing two lists. I tried to understand it but it went over my head.

here is not working example #1

f = Permutations[Range[5], {3}]
DeleteCases[f, CoprimeQ]

here is not working example #2

f = Permutations[Range[5], {3}]
DeleteCases[f, (CoprimeQ[_x] == False) -> x]

also I prefer not to use shorthand @@ @/ @@@. you can see I use the full form of Apply[] which is preferable to me while I am still learning. I can convert between them if I need to so post what you have.

Answer 1

DeleteCases[{x__ /; Not@CoprimeQ[x]}]@f

DeleteCases[Except[_List?(Apply[CoprimeQ])]]@f

both give

{{1, 2, 3}, {1, 2, 5}, {1, 3, 2}, {1, 3, 4}, {1, 3, 5}, {1, 4, 3}, {1, 4, 5},
{1, 5, 2}, {1, 5, 3}, {1, 5, 4}, {2, 1, 3}, {2, 1, 5}, {2, 3, 1}, {2, 3, 5}, 
{2, 5, 1}, {2, 5, 3},  {3, 1, 2}, {3, 1, 4}, {3, 1, 5}, {3, 2, 1}, {3, 2, 5}, 
{3, 4, 1}, {3, 4, 5}, {3, 5, 1},  {3, 5, 2}, {3, 5, 4}, {4, 1, 3}, {4, 1, 5}, 
{4, 3, 1}, {4, 3, 5}, {4, 5, 1}, {4, 5, 3}, {5, 1, 2}, {5, 1, 3}, {5, 1, 4}, 
{5, 2, 1}, {5, 2, 3}, {5, 3, 1}, {5, 3, 2}, {5, 3, 4}, {5, 4, 1}, {5, 4, 3}}

You an also use

Cases[{x__ /; CoprimeQ[x]}]@f

Cases[_List?(Apply[CoprimeQ])]@f

Select[Apply[CoprimeQ]]@f

Pick[f, Apply[CoprimeQ, f, 2]]

Answer 2

after reading and testing the excellent examples provided by kglr, I found a solution based on this answer that is better suited to my preferred choice of syntax style.

Select[Subsets[Range[5], {3}], Apply[CoprimeQ]]

I was making multiple mistakes. I was using Permutations instead of Subsets. it was difficult to find documentation on @ but I found it here. not exactly the best wolfram documentation for more than one argument but it got me through my misunderstanding of the example posted here. this was adapted from

f = Permutations[Range[5], {3}]
Select[Apply[CoprimeQ]]@f

notice all the @ /@ ? __/; is gone after I reworked it for my needs.

after I consolidated everything to a single line of code using pure functions, I was then able to make a Table. example:

Table[
Length[Select[Subsets[Range[x], {3}], Apply[CoprimeQ]]], {x, 3, 100}
]

which gives

{1, 2, 7, 8, 19, 25, 37, 42, 73, 79, 124, 138, 159, 183, 262, 277,
378, 405, 454, 491, 640, 668, 794, 850, 959, 1016, 1257, 1285, 1562,
1668, 1805, 1905, 2088, 2150, 2545, 2673, 2866, 2968, 3457, 3522,
4063, 4228, 4431, 4620, 5269, 5385, 5936, 6106, 6453, 6684, 7513,
7660, 8166, 8400, 8834, 9141, 10168, 10287, 11388, 11752, 12202,
12614, 13341, 13521, 14848, 15263, 15921, 16164, 17657, 17918, 19505,
20031, 20603, 21125, 22248, 22509, 24364, 24799, 25796, 26434, 28493,
28761, 30057, 30766, 31830, 32476, 34843, 35115, 36728, 37513, 38734,
39577, 41208, 41665, 44470, 45192, 46424, 47086}

I then found a match in OEIS. with code example

a[n_] := Select[Subsets[Range[n], {3}], And @@ (GCD @@ # == 1 & /@ Subsets[#, {2}]) &] // Length;
a /@ Range[49]

you can decide who has the cleanest code out of all the examples. big thanks to kglr for helping me through when I was stuck.