0 0
1 30
2 60
3 90
4 120
5 150
6 180
7 210
8 240
9 270
10 300
11 330
hello,
i think it's (pseudo-code)
array 1 : 0..11
array Degrees : 0..330
then two fors, one in another
for i ( 0..11
for j ( 0..11
println array 1 [i]
println array Degrees [j]
DoSomethingWith (array 1 [i] ,array Degrees [j] )
}
}
hope this helps
9 09 309 609 909 1209 1509 1809 2109 2409 2709 30010 010 3010 6010 9010 12010 15010 18010 21010 24010 27010 300
I think you want two things:
1.
choose three numbers in every combination from your set (30-60-90, but also 30-300-180 and so forth)
2.
have each of those three numbers in every possible order (30-60-90, 90-60-30 and so forth).
I think that is *far* more than 144…
Approach:
Three fors within each other (pseudo-code)
for i {
for j {
for k {
if (i!=k && i!=j && j!=k) {
haveOrderChanged(degr[i], degr[j], degr[k]);
}
}
}
}
doSomethingWith (degr1, degr3, degr2);
doSomethingWith (degr2, degr1, degr3);
doSomethingWith (degr2, degr3, degr1);
doSomethingWith (degr3, degr1, degr2);
doSomethingWith (degr3, degr2, degr1);
}
hope, this helps.
Greetings!
0 30 60 900 30 90 600 60 30 900 60 90 300 90 30 600 90 60 3030 0 60 9030 0 90 6030 60 0 9030 60 90 030 90 0 6030 90 60 060 0 30 9060 0 90 3060 30 0 9060 30 90 060 90 0 3060 90 30 090 0 30 6090 0 60 3090 30 0 6090 30 60 090 60 0 3090 60 30 0
I'm not really interested in the possible amount of permutions, i would like to get the permutions (like in one of the previous posts).For example, suppose we have a set of three letters: A, B, and C. We might ask how many ways we can arrange 2 letters from that set. Each possible arrangement would be an example of a permutation. The complete list of possible permutations would be: AB, AC, BA, BC, CA, and CB.
