Python or math: How to count all possible combinations of a list's elements?
Question:
Say there is a list [1,2,3,4,5], I would need to get the count of all possible combinations of the elements (or ‘sub-lists’), e.g. 1, 2, 3, 4, 5, 12, 13, 14, ..., 123, 124, ..., 12345.
I know how to get nCr, the count of combinations of r elements of a list with total n elements.
Python 3.8 or above:
from math import comb
p, r = 5, 2
print(comb(p, r))
Then I could do nC1 + nC2 +...+ nCn. But is there a better/faster way?
p, result = 5, 0
for r in range(1, 6):
result += comb(p, r)
print(result)
Would appreciate your answers.
Answers:
This specific sum is equal to 2^n -1 🙂
This concept is called a power set in mathematics, and it refers to all subsets of a given set. Your question refers to the size of the power set which is 2^n where n is the size of your original set. This total includes the empty set, so as C4stor said, your total would be 2^n - 1.
Say there is a list [1,2,3,4,5], I would need to get the count of all possible combinations of the elements (or ‘sub-lists’), e.g. 1, 2, 3, 4, 5, 12, 13, 14, ..., 123, 124, ..., 12345.
I know how to get nCr, the count of combinations of r elements of a list with total n elements.
Python 3.8 or above:
from math import comb
p, r = 5, 2
print(comb(p, r))
Then I could do nC1 + nC2 +...+ nCn. But is there a better/faster way?
p, result = 5, 0
for r in range(1, 6):
result += comb(p, r)
print(result)
Would appreciate your answers.
This specific sum is equal to 2^n -1 🙂
This concept is called a power set in mathematics, and it refers to all subsets of a given set. Your question refers to the size of the power set which is 2^n where n is the size of your original set. This total includes the empty set, so as C4stor said, your total would be 2^n - 1.