Problem:
Score : 300 points

Problem Statement

We have a grid of H rows and W columns of squares. The color of the square at the i-th row from the top and the j-th column from the left (1 \leq i \leq H, 1 \leq j \leq W) is given to you as a character c_{i,j}: the square is white if c_{i,j} is ., and black if c_{i,j} is #.

Consider doing the following operation:

Choose some number of rows (possibly zero), and some number of columns (possibly zero). Then, paint red all squares in the chosen rows and all squares in the chosen columns.

You are given a positive integer K. How many choices of rows and columns result in exactly K black squares remaining after the operation? Here, we consider two choices different when there is a row or column chosen in only one of those choices.

Constraints

1 \leq H, W \leq 6

1 \leq K \leq HW

c_{i,j} is . or #.

Input

Input is given from Standard Input in the following format:

H W K
c_{1,1}c_{1,2}...c_{1,W}
c_{2,1}c_{2,2}...c_{2,W}
:
c_{H,1}c_{H,2}...c_{H,W}

Output

Print an integer representing the number of choices of rows and columns satisfying the condition.

Sample Input 1

2 3 2
..#
###

Sample Output 1

5

Five choices below satisfy the condition.

The 1-st row and 1-st column

The 1-st row and 2-nd column

The 1-st row and 3-rd column

The 1-st and 2-nd column

The 3-rd column

Sample Input 2

2 3 4
..#
###

Sample Output 2

1

One choice, which is choosing nothing, satisfies the condition.

Sample Input 3

2 2 3
##
##

Sample Output 3

0

Sample Input 4

6 6 8
..##..
.#..#.
#....#
######
#....#
#....#

Sample Output 4

208

Solution in Rust (1.42.0) (judge verdict: Accepted)

#[allow(unused_imports)]
#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::HashMap;
#[allow(unused_imports)]
use std::f64::consts::PI;
#[allow(unused_imports)]
use std::fmt;
#[allow(unused_imports)]
use std::fmt::Debug;
#[allow(unused_imports)]
use std::io;
#[allow(unused_imports)]
use std::str::FromStr;

fn main() {
    // setup
    let input = get_vec_input();
    let h: usize = input[0];
    let w = input[1];
    let k = input[2];

    let table = (0..h)
        .map(|_| get_string_input())
        .map(|x| x.chars().map(|x| x.to_string()).collect::<Vec<_>>())
        .collect::<Vec<Vec<_>>>();

    let binary_string = get_binary_string(h + w);

    // exercise
    let mut answer = 0;
    for b in binary_string {
        let mut count = 0;
        for i in 0..h {
            for j in 0..w {
                if table[i][j] == "#".to_string()
                    && b[i] == '0'.to_string()
                    && b[h + j] == '0'.to_string()
                {
                    count += 1
                }
            }
        }
        if count == k {
            answer += 1;
        }
    }

    // result
    println!("{}", answer);
}

#[allow(dead_code)]
fn get_radian_from(degree: f64) -> f64 {
    (degree * PI) / 180.0
}

#[allow(dead_code)]
fn get_alphabet() -> Vec<char> {
    (0..26).map(|x| (x + b'a') as char).collect::<Vec<char>>()
}

#[allow(dead_code)]
fn get_binary_string(digit: usize) -> Vec<Vec<String>> {
    (0..2i64.pow(digit as u32))
        .map(|x| format!("{:0>1$b}", x, digit))
        .map(|x| x.chars().map(|x| x.to_string()).collect::<Vec<String>>())
        .collect::<Vec<Vec<_>>>()
}

#[allow(dead_code)]
fn check_prime(value: i64) -> bool {
    if value == 2 {
        return true;
    }
    if value % 2 == 0 {
        return false;
    }
    for i in 3..(value as f64).sqrt().floor() as i64 {
        if value % i == 0 {
            return false;
        }
    }
    return true;
}

#[allow(dead_code)]
fn get_digit(value: i64) -> i64 {
    let mut digit = 0i64;
    let mut value = value;
    while value != 0 {
        value /= 10;
        digit += 1;
    }
    digit
}

#[allow(dead_code)]
fn sum_digit2(value: String) -> u32 {
    return value.chars().fold(0u32, |a, b| a + b.to_digit(10).unwrap());
}

#[allow(dead_code)]
fn sum_digit(value: i64) -> i64 {
    if value < 10 {
        return value;
    }
    return sum_digit(value / 10) + value % 10;
}

#[allow(dead_code)]
fn update_max<T>(left: &mut T, right: T)
where
    T: std::cmp::Ord + std::clone::Clone,
{
    if *left < right {
        *left = right.clone();
    }
}

#[allow(dead_code)]
fn update_min<T>(left: &mut T, right: T)
where
    T: std::cmp::Ord + std::clone::Clone,
{
    if *left > right {
        *left = right.clone();
    }
}

#[allow(dead_code)]
fn get_vec_input<T>() -> Vec<T>
where
    T: FromStr,
    T::Err: Debug,
{
    let mut input = String::new();
    io::stdin()
        .read_line(&mut input)
        .expect("Failed to read line");
    let vec: Vec<T> = input
        .split_whitespace()
        .map(|x| x.parse().unwrap())
        .collect();
    vec
}

#[allow(dead_code)]
fn get_tuple_input() -> (u64, f32) {
    let mut s = String::new();
    io::stdin().read_line(&mut s).expect("");
    let mut iter = s.trim().split_whitespace();
    let c: u64 = iter.next().unwrap().parse().unwrap();
    let n: f32 = iter.next().unwrap().parse().unwrap();
    (c, n)
}

#[allow(dead_code)]
fn get_string_input() -> String {
    let mut s = String::new();
    io::stdin().read_line(&mut s).expect("");
    s.trim().to_string()
}

/// https://github.com/bluss/permutohedron/blob/master/src/lexical.rs
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;
}

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
    }
}