Given the digits 1, 2, 3, and 4, how many distinct three-digit numbers can be formed without repeating any digit? This is a classic combinatorial problem equivalent to calculating permutations of 4 items taken 3 at a time, denoted as P(4,3) = 4! / (4-3)! = 24.

Brute-force Approach with Nested Loops

A straightforward solution uses three nested loops to enumerate all possilbe combinations, filtering out those with repeated digits.

total = 0
for a in range(1, 5):
    for b in range(1, 5):
        for c in range(1, 5):
            if a != b and a != c and b != c:
                total += 1
                print(f"Combination #{total}: {a},{b},{c}")

Using itertools.permutations

Python's itertools module simplifies this task. We can generate all ordered permutations of length 3 from the set {'1','2','3','4'} and convert each tuple into an integer.

from itertools import permutations

numbers = [int(''.join(x)) for x in permutations('1234', 3)]
print(f"Total numbers: {len(numbers)}")
print(numbers)

This returns a list of 24 integers, each representing a valid three-digit number.

Extension: Handling Repeated Digits

If repetition were alllowed, we would use itertools.product to generate the Cartesian product.

from itertools import product

# With repetition: each position can be 1-4
numbers_with_rep = [int(''.join(x)) for x in product('1234', repeat=3)]
print(f"Total with repetition: {len(numbers_with_rep)}")

This produces 4^3 = 64 numbers.

Summary of itertools Combinatorial Iterators

Iterator	Arguments	Description	Example
product(p, q, ..., repeat=1)	p, q, ... [repeat=1]	Cartesian product, equivalent to nested for loops	product('ABCD', repeat=2) gives AA, AB, ... DD
permutations(p[, r])	p[, r]	r-length tuples, all possible orderings, no repeated elements	permutations('ABCD', 2) gives AB, AC, ... DC
combinations(p, r)	p, r	r-length tuples in sorted order, no repeated elements	combinations('ABCD', 2) gives AB, AC, AD, BC, BD, CD
combinations_with_replacement(p, r)	p, r	r-langth tuples in sorted order, with repeated elements allowed	combinations_with_replacement('ABCD',2) gives AA, AB, ... DD