Select random permutations and save the values in different variables

Question

When I create the permutations below, there are 360 possibilities.

#Print permutations of 1 to 6 of a size 4
from itertools import permutations


possibilities = permutations([1,2,3,4,5,6], 4) #Permutations of size 6

sum_of_possibilities = 0

for i in list(possibilities):
    print(i)
    sum_of_possibilities = sum_of_possibilities +1

I would like to get 4 random permutations of these 360 and save each number in a different variable.

For examples:

Select these 4 permutations below

(3, 5, 6, 2)
(1, 6, 4, 3)
(4, 3, 1, 5)
(6, 5, 3, 4)

and save the first number 3 as x1; 5 as x2; 6 as x3; 2 as x4... 3 as x15 and 4 as x16.

#RESULTS

print(f'Total of possibilities are: {sum_of_possibilities}')

print(possibilities[0])

Total of possibilities are: 360

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
~\AppData\Local\Temp\ipykernel_13912\1726349991.py in <module>
      3 print(f'Total of possibilities are: {sum_of_possibilities}')
      4 
----> 5 print(possibilities[0])

TypeError: 'itertools.permutations' object is not subscriptable

As a side note, generating all possible permutations becomes very costly very quickly for increasing population sizes. It might be worth consider whether this is really necessary, or if a simpler approach would work just as well, like picking random samples without replacement — Brian61354270
– Brian61354270, Commented Feb 7, 2023 at 20:56

Lucas · Accepted Answer · 2023-02-08 00:02:40Z

Got it!

#Print permutations of 1 to 6 of a size 6
from itertools import permutations #import library to do permutations

size = 6

possibilities = list(permutations([1,2,3,4,5,6], size)) #Permutations of size 6

print(f'There are {len(possibilities)} possibilities to be analyzed\n')

print('The solutions are (i0, i1, i2, i3, i4, i5) = ')

solutions=0 #quantity of solutions found before start to calculate

for n in range(len(possibilities)):
    i=[]
    for j in range(size):
        i.append(possibilities[n][j])
        
    c1 = i[0]+i[2]+i[3]+i[4]
    c2 = i[0]+i[1]+i[3]+i[5]
    c3 = i[4]+i[1]+i[2]+i[5]
    if c1 == c2 and c1 == c3: #Test if the sums are the same
        sum = c1
        print(possibilities[n])
        solutions = solutions + 1
print(f'\n There are {solutions} possible solutions for this puzzle, where all the sum in the circle are equal to {sum}')

It's a math puzzle https://github.com/lstavares84/math_puzzles_in_python/blob/main/math_puzzle_1.ipynb

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