Math + combinatorics [EXPLAINED]
IntuitionI noticed that the problem is about finding how many ways we can rearrange the letters of each word while keeping the words in their original positions
# Intuition
I noticed that the problem is about finding how many ways we can rearrange the letters of each word while keeping the words in their original positions. Since we only care about permutations of the letters in each word, I focused on counting all unique arrangements for every word separately.

# Approach
Break the sentence into individual words, compute how many unique ways each word's letters can be rearranged by using the counts of each letter to adjust the total number of possibilities, and then combine the results for all the words while keeping everything within a set mathematical limit to handle large results.

# Complexity
- Time complexity:
O(n + m) where n is the length of the string for splitting and m is for precomputing factorials.

- Space complexity:
O(m) for storing factorials and character counts.

# Code
```rust []
impl Solution {
    pub fn count_anagrams(s: String) -> i32 {
        const MOD: i64 = 1_000_000_007;

        // Helper function to compute factorial modulo MOD
        fn factorial_mod(n: usize, mod_val: i64) -> Vec<i64> {
            let mut fact = vec![1; n + 1];
            for i in 2..=n {
                fact[i] = fact[i - 1] * i as i64 % mod_val;
            }
            fact
        }

        // Helper function to compute modular multiplicative inverse
        fn mod_inverse(x: i64, mod_val: i64) -> i64 {
            let mut base = x;
            let mut exp = mod_val - 2;
            let mut result = 1;
            while exp > 0 {
                if exp % 2 == 1 {
                    result = result * base % mod_val;
                }
                base = base * base % mod_val;
                exp /= 2;
            }
            result
        }

        // Precompute factorials and modular inverses up to the max possible length
        let max_len = 100_000;
        let fact = factorial_mod(max_len, MOD);

        s.split_whitespace()
            .map(|word| {
                let mut freq = vec![0; 26];
                for &b in word.as_bytes() {
                    freq[(b - b'a') as usize] += 1;
                }

                let mut result = fact[word.len()];
                for &count in &freq {
                    if count > 1 {
                        result = result * mod_inverse(fact[count], MOD) % MOD;
                    }
                }
                result
            })
            .fold(1, |acc, x| acc * x % MOD) as i32
    }
}

```
['Hash Table', 'Math', 'String', 'Combinatorics', 'Counting', 'Rust']