use std::vec::Vec;
use std::cmp::*;

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace().map(|e| e.parse().ok().unwrap()).collect()
}


/// Permute a slice into its next or previous permutation (in lexical order).
///
/// ```
/// use permutohedron::LexicalPermutation;
///
/// let mut data = [1, 2, 3];
/// let mut permutations = Vec::new();
///
/// loop {
///     permutations.push(data.to_vec());
///     if !data.next_permutation() {
///         break;
///     }
/// }
///
/// assert_eq!(permutations, &[&[1, 2, 3], &[1, 3, 2],
///                            &[2, 1, 3], &[2, 3, 1],
///                            &[3, 1, 2], &[3, 2, 1]]);
///
/// // `data` has been mutated in-place:
/// assert_eq!(data, [3, 2, 1]);
/// ```
pub trait LexicalPermutation {
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the last ordered permutation.
    fn next_permutation(&mut self) -> bool;
    /// Return `true` if the slice was permuted, `false` if it is already
    /// at the first ordered permutation.
    fn prev_permutation(&mut self) -> bool;
}

impl<T> LexicalPermutation for [T] where T: Ord {
    /// Original author in Rust: Thomas Backman <serenity@exscape.org>
    fn next_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }

        // Step 1: Identify the longest, rightmost weakly decreasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] >= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the last-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Find the rightmost element larger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j] <= self[i-1]  {
            j -= 1;
        }

        // Step 3: Swap that element with the pivot
        self.swap(j, i-1);

        // Step 4: Reverse the (previously) weakly decreasing part
        self[i..].reverse();

        true
    }

    fn prev_permutation(&mut self) -> bool {
        // These cases only have 1 permutation each, so we can't do anything.
        if self.len() < 2 { return false; }

        // Step 1: Identify the longest, rightmost weakly increasing part of the vector
        let mut i = self.len() - 1;
        while i > 0 && self[i-1] <= self[i] {
            i -= 1;
        }

        // If that is the entire vector, this is the first-ordered permutation.
        if i == 0 {
            return false;
        }

        // Step 2: Reverse the weakly increasing part
        self[i..].reverse();

        // Step 3: Find the rightmost element equal to or bigger than the pivot (i-1)
        let mut j = self.len() - 1;
        while j >= i && self[j-1] < self[i-1]  {
            j -= 1;
        }

        // Step 4: Swap that element with the pivot
        self.swap(i-1, j);

        true
    }

}


fn main() {
  let n: usize = read();

  let mut v: Vec<u64> = vec![0; 5];
  for _ in 0..n {
    let s: Vec<char> = read::<String>().chars().collect();
    if s[0] == 'M' {
      v[0] += 1;
    }
    else if s[0] == 'A' {
      v[1] += 1;
    }
    else if s[0] == 'R' {
      v[2] += 1;
    }
    else if s[0] == 'C' {
      v[3] += 1;
    }
    else if s[0] == 'H' {
      v[4] += 1;
    }
  }

  let mut data = vec![0, 0, 1, 1, 1];
  let mut res = 0;
  loop {
    let mut count = 1;
    for i in 0..5 {
      if data[i] == 1 {
        count *= v[i];
      }
    }
    res += count;
    if !data.next_permutation() {
      break;
    }
  }

  println!("{}", res);
}


Measured memory_bytes: 4352