How to form all possible combination of given digits to form a n-digit number without repetition of digits?(python)

Question

Question

Let say we have a list of given digits and n (to form n-digit number)

digits = [0, 1, 2]
n = 3

List digit can not contain duplicate values.

How we can form all possible n digit numbers without repetition of digits. n can be any positive integer and list digits can have any length and digits in it.

If length of list digits is less then n then no valid numbers can be formed.

---

Conditions on number

Valid Numbers

[120, 102, 201, 210]

Invalis Numbers

[012, 112, 221, 212, 121]

012 is invalid because it is equivalent to 12 which is not a 3-digit number

---

Here is what I tried

def formNum(L):
    num = []
    for i in range(3):
        for j in range(3):
            for k in range(3):
                  
                if (i!=j and i != 0 and j!=k and i!=k):
                    num.append(int(f"{L[i]}{L[j]}{L[k]}"))
                      
    return num

print(formNum([0, 1, 2]))

Output

[102, 120, 201, 210]

But my solution will only work for 3 digits number and is very difficult to expand and for large value of n it is very slow.

---

My question is not similar to this one as this question ask for all possible arrangements of given digits but mine ask for all possible arrangements of digit to form a number of n-digits (n is given)

Answer 1

This is a possible solution:

from itertools import permutations

num = [int(''.join(map(str, p))) for p in permutations(digits, n) if p[0] != 0]

The output for your example:

[102, 120, 201, 210]