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 C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>   //PBDS
#include <ext/pb_ds/tree_policy.hpp>       //PBDS
#define int long long int
#define vi vector<int>
#define vii vector<pair<int,int>>
#define pi pair<int,int>
#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define mi map<int,int>
#define PI   3.141592653589
#define all(a)  (a).begin(),(a).end()
#define rall(a) (a).rbegin(),(a).rend()
#define f(i,x,n) for(int i=x;i<n;i++)
#define rf(i,x,n) for(int i=n-1;i>=x;i--)
#define check(x)   cerr<<#x<<" : "<<x<<endl;
#define ordered_set tree<int,null_type,less_equal<int>,rb_tree_tag,tree_order_statistics_node_update>
#define fbo find_by_order                    //return iterator PBDS
#define ook order_of_key                     //return number :)) PBDS
#define baar exit(0)
using namespace __gnu_pbds;
using namespace std;
int powmod(int base,int exponent,int mod)
{
	int ans=1;
	while(exponent){
		if(exponent&1)ans=(ans*base)%mod;
		base=(base*base)%mod;
		exponent/=2;
	}
	return ans;
}
signed main()
{
	fast
	int h,w,k;
	cin>>h>>w>>k;
	string  ans[h];
	int ttl=0;
	f(i,0,h)
	{
		cin>>ans[i];
		f(j,0,w)
		{
			if(ans[i][j]=='#') ttl++;
		}
    }
    int db=0;
    f(i,0,(1LL<<h))
    {
    	f(j,0,(1LL<<w))
    	{
    		vi row,col;
    		int val=ttl;
    		f(k,0,32)
    		{
    			if(i&(1<<k)) row.pb(k);
			}
			f(k,0,32)
			{
				if(j&(1<<k)) col.pb(k);
			}
			int vis[7][7];
			memset(vis,0,sizeof(vis));
			for(int x:row)
			{
				for(int l=0;l<w;l++)
				{
					if(ans[x][l]=='#') 
					{
						val--;
						vis[x][l]=1;
					}
				}
			}
			for(int x:col)
			{
				for(int l=0;l<h;l++)
				{
					if(vis[l][x]==0 && ans[l][x]=='#')
					{
						val--;
					}
				}
			}
			if(val==k) db++;
		}
	}
	cout<<db;
	//cerr << "Time elapsed: " << clock() / 1000 << " ms" << endl;
    return 0;
}