Comprehensive Solution by Reducing Problems Into Familiar Subproblems
IntuitionThis problem is about combinatorics, hence we make use of the math.comb library. We also need to keep a character frequency of each word so that we can
# Intuition
This problem is about combinatorics, hence we make use of the math.comb library. We also need to keep a character frequency of each word so that we can calculate combinations where there are character repetitions

# Problem Decomposition
We can reduce this problem to the following subproblems:
- Converting space-delimited string into an array
- Traversing an array
- Creating and accessing values in a hashmap
- Calculating combinatorics
# Approach
To begin with, lets define a helper function `calculateFactorial` that takes a single `word` and the number of all possible combinations/ arrangements that `word` can take.
Within the scope of `calculateFactorial`:
- we create a hashmap `table` to store the frequency counts of all letters in `word`.
- save the length of word as `n`
- establish a `total` variable to save the answer for this function
- traverse the hashmap `table` and for each iteration:
- - multiply the current `total` by `n`! / (`n`! - `table[`w`]`!). This here is handled by making use of math.comb
- - decrement `n` by `table`[`w`]`. 
- outside the traversal, return `total`

Now. let's implement the main function:
First, we establish a variable `res` that we shall use to incrementally build our final solution.
Split `s` into an array of words called `words`
Traverse the new array `words` and for each iteration:
- we multiply `res` by the combinatorial result of calling `calculateWordFactorial` on the current word. We ensure to bound the result using the provided modulus.
Outside of the for loop, we simply return `res`
# Complexity
- Time complexity:
`countAnagrams` - O(n)
`calculateWordFactorial` - O(k) where `k` == len('word')
total = O(nk)
- Space complexity:
`countAnagrams` - O(n)
`calculateWordFactorial` - O(k) where `k` == len('word')
total = O(nk)

# Code
```python3 []
import math
from collections import Counter

class Solution:
    def countAnagrams(self, s: str) -> int:
        '''
        This is a mathematical problem. We can find the distinct number of permutation of each word in the string and multiply using the combinatorial formular
        '''


        res = 1
        words = s.split(" ")

        for word in words:
            res = (res * self.calculateWordFactorial(word)) % (10**9 + 7)

        return res

    def calculateWordFactorial(self, word):
        table = Counter(word)
        n = len(word)
        total = 1
    
        for w in table:
            total *= math.comb(n, table[w])
            n -= table[w]
            

        return int(total)


        
        
```
['Hash Table', 'String', 'Combinatorics', 'Python3']