freecodecamp--freecodecamp/curriculum/challenges/english/blocks/daily-coding-challenges-python/69162d64f96574d9bb629f01.md

id, title, challengeType, dashedName

id	title	challengeType	dashedName
69162d64f96574d9bb629f01	Challenge 116: Permutation Count	29	challenge-116

--description--

Given a string, return the number of distinct permutations that can be formed from its characters.

• A permutation is any reordering of the characters in the string.
• Do not count duplicate permutations.
• If the string contains repeated characters, repeated arrangements should only be counted once.
• The string will contain only letters (A-Z, a-z).

For example, given "abb", return 3 because there's three unique ways to arrange the letters: "abb", "bab", and "bba".

--hints--

count_permutations("abb") should return 3.

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_permutations("abb"), 3)`)
}})

count_permutations("abc") should return 6.

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_permutations("abc"), 6)`)
}})

count_permutations("racecar") should return 630.

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_permutations("racecar"), 630)`)
}})

count_permutations("freecodecamp") should return 39916800.

({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_permutations("freecodecamp"), 39916800)`)
}})

--seed--

--seed-contents--

def count_permutations(s):

    return s

--solutions--

from math import factorial
from collections import Counter
def count_permutations(s):
    freq = Counter(s)
    n = len(s)

    result = factorial(n)
    for count in freq.values():
        result //= factorial(count)

    return result