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 Go (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
	"strings"
)

const (
	initialBufSize = 1e4
	maxBufSize     = 1e6 + 7
)

var buf []byte = make([]byte, maxBufSize)
var sc = bufio.NewScanner(os.Stdin)

func init() {
	sc.Split(bufio.ScanWords)
	sc.Buffer(buf, maxBufSize)
}

// bit全探索

func main() {
	H, W, K := nextInt(), nextInt(), nextInt()
	Chw := make([][]string, H)
	for i := 0; i < H; i++ {
		c := nextLine()
		Chw[i] = strings.Split(c, "")
	}
	var ans int
	for maskR := 0; maskR < 1<<H; maskR++ {
		for maskC := 0; maskC < 1<<W; maskC++ {
			var black int
			for i := 0; i < H; i++ {
				for j := 0; j < W; j++ {
					if (maskR>>i)&1 == 0 && (maskC>>j)&1 == 0 && Chw[i][j] == "#" {
						black++
					}
				}
			}
			if black == K {
				ans++
			}
		}
	}
	fmt.Println(ans)
}

func nextLine() string {
	sc.Scan()
	return sc.Text()
}

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func abs(a int) int {
	if a > 0 {
		return a
	} else {
		return -a
	}
}

func pow(p, q int) int {
	res := p
	if q == 0 {
		return 1
	}
	for i := 0; i < q-1; i++ {
		res *= p
	}
	return res
}

func MinOf(vars ...int) int {
	min := vars[0]
	for _, i := range vars {
		if min > i {
			min = i
		}
	}
	return min
}

func MaxOf(vars ...int) int {
	max := vars[0]
	for _, i := range vars {
		if max < i {
			max = i
		}
	}
	return max
}

func gcdof2numbers(a int, b int) int {
	if b == 0 {
		return a
	}
	return gcdof2numbers(b, a%b)
}

func lcmof2numbers(a int, b int) int {
	return a / gcdof2numbers(a, b) * b
}

// マイナスの場合は考慮していない
func fuctorial(a int) int {
	if a == 1 || a == 0 {
		return 1
	} else {
		return fuctorial(a-1) * a
	}
}

func NextPermutation(x sort.Interface) bool {
	n := x.Len() - 1
	if n < 1 {
		return false
	}
	j := n - 1
	for ; !x.Less(j, j+1); j-- {
		if j == 0 {
			return false
		}
	}
	l := n
	for !x.Less(j, l) {
		l--
	}
	x.Swap(j, l)
	for k, l := j+1, n; k < l; {
		x.Swap(k, l)
		k++
		l--
	}
	return true
}

// 素因数分解したmapを返す
func primeFuctorize(n int) map[int]int {
	pf := map[int]int{}
	for i := 2; i*i <= n; i++ {
		for n%i == 0 {
			pf[i]++
			n /= i
		}
	}
	if n != 1 {
		pf[n]++
	}
	return pf
}

// ここから拝借
// https://shoman.hatenablog.com/entry/2020/02/25/185456
// Stackは[]intのエイリアス
type Stack []int

// Push adds an element
func (s *Stack) Push(v int) {
	*s = append(*s, v)
}

// Pop removes the top element and return it
func (s *Stack) Pop() (int, error) {
	if s.Empty() {
		return 0, fmt.Errorf("stack is empty")
	}

	v := (*s)[len(*s)-1]
	*s = (*s)[:len(*s)-1]
	return v, nil
}

// Peek returns the top value
func (s *Stack) Peek() (int, error) {
	if s.Empty() {
		return 0, fmt.Errorf("stack is empty")
	}

	return (*s)[len(*s)-1], nil
}

// Size returns the length of stack
func (s *Stack) Size() int {
	return len(*s)
}

// Empty returns true when stack is empty
func (s *Stack) Empty() bool {
	return s.Size() == 0
}

// NewStack generates stack
func NewStack() *Stack {
	s := new(Stack)
	return s
}