Given integers N, P, and Q the task is to find the number of ways to form a group of N people having at least 4 boys and 1 girl from P boys and Q girls.
Examples:
Input: P = 5, Q = 2, N = 5
Output: 10
Explanation: Suppose given pool is {m1, m2, m3, m4, m5} and {w1, w2}. Then possible combinations are:
m1 m2 m3 m4 w1
m2 m3 m4 m5 w1
m1 m3 m4 m5 w1
m1 m2 m4 m5 w1
m1 m2 m3 m5 w1
m1 m2 m3 m4 w2
m2 m3 m4 m5 w2
m1 m3 m4 m5 w2
m1 m2 m4 m5 w2
m1 m2 m3 m5 w2Hence the count is 10.
Input: P = 5, Q = 2, N = 6
Output: 7
Naive Approach: This problem is based on combinatorics, and details of the Naive approach is already discussed in Set-1 of this problem.
For some general value of P, Q and N we can calculate the total possible ways using the following formula:
where
In this approach at every step we were calculating the value for each possible way.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To solve this problem efficiently, we can use the Pascal Triangle property to calculate the
1
1 1
1 2 1
1 3 3 1
.
.
.
which is nothing but
.
.
.
Follow the steps mentioned below:
- Use the pascal triangle to precalculate the values of the combination.
- Start iterating a loop from i = 4 to i = P and do the following for each iteration.
- Check if (N-i) ≥ 1 and (N-i) ≤ Q.
- If the condition is satisfied then count the possible ways for i men and (N-i) women, otherwise, skip the step.
- Add the count with the total number of ways.
- Return the total count as your answer.
Below is the implementation of the approach:
#include <bits/stdc++.h> using namespace std;
long long int pascal[31][31];
// Function to calculate the pascal triangle void pascalTriangle()
{ pascal[0][0] = 1;
pascal[1][0] = 1;
pascal[1][1] = 1;
// Loop to calculate values of
// pascal triangle
for ( int i = 2; i < 31; i++) {
pascal[i][0] = 1;
for ( int j = 1; j < i; j++)
pascal[i][j]
= pascal[i - 1][j]
+ pascal[i - 1][j - 1];
pascal[i][i] = 1;
}
} // Function to calculate the number of ways long long int countWays( int n, int p, int q)
{ // Variable to store the answer
long long int sum = 0;
// Loop to calculate the number of ways
for ( long long int i = 4; i <= p; i++) {
if (n - i >= 1 && n - i <= q)
sum += pascal[p][i]
* pascal[q][n - i];
}
return sum;
} // Driver code int main()
{ pascalTriangle();
int P = 5, Q = 2, N = 5;
// Calculate possible ways for given
// N, P, and Q
cout << countWays(N, P, Q) << endl;
return 0;
} |
import java.util.*;
public class GFG
{ static long [][]pascal = new long [ 31 ][ 31 ];
// Function to calculate the pascal triangle static void pascalTriangle()
{ pascal[ 0 ][ 0 ] = 1 ;
pascal[ 1 ][ 0 ] = 1 ;
pascal[ 1 ][ 1 ] = 1 ;
// Loop to calculate values of
// pascal triangle
for ( int i = 2 ; i < 31 ; i++) {
pascal[i][ 0 ] = ( int ) 1 ;
for ( int j = 1 ; j < i; j++)
pascal[i][j]
= pascal[i - 1 ][j]
+ pascal[i - 1 ][j - 1 ];
pascal[i][i] = 1 ;
}
} // Function to calculate the number of ways static long countWays( int n, int p, int q)
{ // Variable to store the answer
long sum = 0 ;
// Loop to calculate the number of ways
for ( int i = 4 ; i <= p; i++) {
if (n - i >= 1 && n - i <= q) {
sum += ( int )pascal[p][i]
* ( int )pascal[q][n - i];
}
}
return sum;
} // Driver code public static void main(String args[])
{ pascalTriangle();
int P = 5 , Q = 2 , N = 5 ;
// Calculate possible ways for given
// N, P, and Q
System.out.print(countWays(N, P, Q));
} } // This code is contributed by Samim Hossain Mondal. |
# Python3 program for the above approach import numpy as np
pascal = np.zeros(( 31 , 31 ));
# Function to calculate the pascal triangle def pascalTriangle() :
pascal[ 0 ][ 0 ] = 1 ;
pascal[ 1 ][ 0 ] = 1 ;
pascal[ 1 ][ 1 ] = 1 ;
# Loop to calculate values of
# pascal triangle
for i in range ( 2 , 31 ) :
pascal[i][ 0 ] = 1 ;
for j in range ( 1 , i) :
pascal[i][j] = pascal[i - 1 ][j] + pascal[i - 1 ][j - 1 ];
pascal[i][i] = 1 ;
# Function to calculate the number of ways def countWays(n, p, q) :
# Variable to store the answer
sum = 0 ;
# Loop to calculate the number of ways
for i in range ( 4 , p + 1 ) :
if (n - i > = 1 and n - i < = q) :
sum + = pascal[p][i] * pascal[q][n - i];
return int ( sum );
# Driver code if __name__ = = "__main__" :
pascalTriangle();
P = 5 ; Q = 2 ; N = 5 ;
# Calculate possible ways for given
# N, P, and Q
print (countWays(N, P, Q));
# This code is contributed by AnkThon
|
using System;
class GFG
{ static long [,]pascal = new long [31, 31];
// Function to calculate the pascal triangle static void pascalTriangle()
{ pascal[0, 0] = 1;
pascal[1, 0] = 1;
pascal[1, 1] = 1;
// Loop to calculate values of
// pascal triangle
for ( int i = 2; i < 31; i++) {
pascal[i, 0] = ( int )1;
for ( int j = 1; j < i; j++)
pascal[i, j]
= pascal[i - 1, j]
+ pascal[i - 1, j - 1];
pascal[i, i] = 1;
}
} // Function to calculate the number of ways static long countWays( int n, int p, int q)
{ // Variable to store the answer
long sum = 0;
// Loop to calculate the number of ways
for ( int i = 4; i <= p; i++) {
if (n - i >= 1 && n - i <= q) {
sum += ( int )pascal[p, i]
* ( int )pascal[q, n - i];
}
}
return sum;
} // Driver code public static void Main()
{ pascalTriangle();
int P = 5, Q = 2, N = 5;
// Calculate possible ways for given
// N, P, and Q
Console.Write(countWays(N, P, Q));
} } // This code is contributed by Samim Hossain Mondal. |
<script> let pascal = new Array(31).fill(0).map(() => new Array(31).fill(0));
// Function to calculate the pascal triangle
const pascalTriangle = () => {
pascal[0][0] = 1;
pascal[1][0] = 1;
pascal[1][1] = 1;
// Loop to calculate values of
// pascal triangle
for (let i = 2; i < 31; i++) {
pascal[i][0] = 1;
for (let j = 1; j < i; j++)
pascal[i][j]
= pascal[i - 1][j]
+ pascal[i - 1][j - 1];
pascal[i][i] = 1;
}
}
// Function to calculate the number of ways
const countWays = (n, p, q) => {
// Variable to store the answer
let sum = 0;
// Loop to calculate the number of ways
for (let i = 4; i <= p; i++) {
if (n - i >= 1 && n - i <= q)
sum += pascal[p][i]
* pascal[q][n - i];
}
return sum;
}
// Driver code
pascalTriangle();
let P = 5, Q = 2, N = 5;
// Calculate possible ways for given
// N, P, and Q
document.write(countWays(N, P, Q));
// This code is contributed by rakeshsahni </script> |
Output
10
Time Complexity: O(N)
Auxiliary Space: O(N2)