Given three numbers(x, y and z) which denote the number of people in the first group, the second group, and third group. We can form groups by selecting people from the first group, the second group and third group such that the following conditions are not void.
- A minimum of one people has to be selected from every group.
- The number of people selected from the first group has to be at least one more than the number of people selected from the third group.
The task is to find the number of ways of forming distinct groups.
Examples:
Input: x = 3, y = 2, z = 1
Output: 9
Lets say x has people (a, b, c)
Y has people (d, e)
Z has people (f)
Then the 9 ways are {a, b, d, f}, {a, b, e, f}, {a, c, d, f}, {a, c, e, f},
{b, c, d, f}, {b, c, e, f}, {a, b, c, d, f}, {a, b, c, e, f} and
{a, b, c, d, e, f}
Input: x = 4, y = 2, z = 1
Output: 27
The problem can be solved using combinatorics. There are three positions(in terms of people from different groups) which need to be filled. The first has to be filled with a number with one or greater than the second position. The third can be filled with any number. We know if we need to fill k positions with N people, then the number of ways of doing it is
- The second position can be filled with i = 1 to i = y people.
- The first position can be filled with j = i+1 to j = x people.
- The third position can be filled with any number of k = 1 to k = z people.
- Hence the common thing is filling the third position with k people. Hence we can take that portion as common.
- Run two loops( i and j) for filling up the second position and first position respectively.
- The number of ways of filling up the positions is
* . - After calculation of all ways to fill up those two positions, we can simply multiply summation of
+ + … since that was the common part in both.
Below is the implementation of the above approach.
// C++ program to find the number of // ways to form the group of people #include <bits/stdc++.h> using namespace std;
int C[1000][1000];
// Function to pre-compute the // Combination using DP void binomialCoeff( int n)
{ int i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++) {
for (j = 0; j <= i; j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
}
// return C[n][k];
} // Function to find the number of ways int numberOfWays( int x, int y, int z)
{ // Function to pre-compute
binomialCoeff(max(x, max(y, z)));
// Sum the Zci
int sum = 0;
for ( int i = 1; i <= z; i++) {
sum = (sum + C[z][i]);
}
// Iterate for second position
int sum1 = 0;
for ( int i = 1; i <= y; i++) {
// Iterate for first position
for ( int j = i + 1; j <= x; j++) {
sum1 = (sum1 + (C[y][i] * C[x][j]));
}
}
// Multiply the common Combination value
sum1 = (sum * sum1);
return sum1;
} // Driver Code int main()
{ int x = 3;
int y = 2;
int z = 1;
cout << numberOfWays(x, y, z);
return 0;
} |
// Java program to find the number of // ways to form the group of peopls class GFG
{ static int C[][] = new int [ 1000 ][ 1000 ];
// Function to pre-compute the // Combination using DP static void binomialCoeff( int n)
{ int i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0 ; i <= n; i++)
{
for (j = 0 ; j <= i; j++)
{
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1 ;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1 ][j - 1 ] + C[i - 1 ][j];
}
}
// return C[n][k];
} // Function to find the number of ways static int numberOfWays( int x, int y, int z)
{ // Function to pre-compute
binomialCoeff(Math.max(x, Math.max(y, z)));
// Sum the Zci
int sum = 0 ;
for ( int i = 1 ; i <= z; i++)
{
sum = (sum + C[z][i]);
}
// Iterate for second position
int sum1 = 0 ;
for ( int i = 1 ; i <= y; i++)
{
// Iterate for first position
for ( int j = i + 1 ; j <= x; j++)
{
sum1 = (sum1 + (C[y][i] * C[x][j]));
}
}
// Multiply the common Combination value
sum1 = (sum * sum1);
return sum1;
} // Driver Code public static void main(String args[])
{ int x = 3 ;
int y = 2 ;
int z = 1 ;
System.out.println(numberOfWays(x, y, z));
} } // This code is contributed by Arnab Kundu |
# Python3 program to find the number of # ways to form the group of peopls C = [[ 0 for i in range ( 1000 )]
for i in range ( 1000 )]
# Function to pre-compute the # Combination using DP def binomialCoeff(n):
i, j = 0 , 0
# Calculate value of Binomial Coefficient
# in bottom up manner
for i in range (n + 1 ):
for j in range (i + 1 ):
# Base Cases
if (j = = 0 or j = = i):
C[i][j] = 1
# Calculate value using previously
# stored values
else :
C[i][j] = C[i - 1 ][j - 1 ] + \
C[i - 1 ][j]
# return C[n][k]
# Function to find the number of ways def numberOfWays(x, y, z):
# Function to pre-compute
binomialCoeff( max (x, max (y, z)))
# Sum the Zci
sum = 0
for i in range ( 1 , z + 1 ):
sum = ( sum + C[z][i])
# Iterate for second position
sum1 = 0
for i in range ( 1 , y + 1 ):
# Iterate for first position
for j in range (i + 1 , x + 1 ):
sum1 = (sum1 + (C[y][i] * C[x][j]))
# Multiply the common Combination value
sum1 = ( sum * sum1)
return sum1
# Driver Code x = 3
y = 2
z = 1
print (numberOfWays(x, y, z))
# This code is contributed by Mohit Kumar |
// C# program to find the number of // ways to form the group of peopls using System;
class GFG
{ static int [,]C = new int [1000,1000];
// Function to pre-compute the // Combination using DP static void binomialCoeff( int n)
{ int i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++)
{
for (j = 0; j <= i; j++)
{
// Base Cases
if (j == 0 || j == i)
C[i,j] = 1;
// Calculate value using previously
// stored values
else
C[i,j] = C[i - 1,j - 1] + C[i - 1,j];
}
}
// return C[n,k];
} // Function to find the number of ways static int numberOfWays( int x, int y, int z)
{ // Function to pre-compute
binomialCoeff(Math.Max(x, Math.Max(y, z)));
// Sum the Zci
int sum = 0;
for ( int i = 1; i <= z; i++)
{
sum = (sum + C[z,i]);
}
// Iterate for second position
int sum1 = 0;
for ( int i = 1; i <= y; i++)
{
// Iterate for first position
for ( int j = i + 1; j <= x; j++)
{
sum1 = (sum1 + (C[y,i] * C[x,j]));
}
}
// Multiply the common Combination value
sum1 = (sum * sum1);
return sum1;
} // Driver Code public static void Main(String []args)
{ int x = 3;
int y = 2;
int z = 1;
Console.WriteLine(numberOfWays(x, y, z));
} } // This code is contributed by Rajput-Ji |
<script> // javascript program to find the number of // ways to form the group of peopls var C = Array(1000).fill().map(()=>Array(1000).fill(0));
// Function to pre-compute the
// Combination using DP
function binomialCoeff(n) {
var i, j;
// Calculate value of Binomial Coefficient
// in bottom up manner
for (i = 0; i <= n; i++) {
for (j = 0; j <= i; j++) {
// Base Cases
if (j == 0 || j == i)
C[i][j] = 1;
// Calculate value using previously
// stored values
else
C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
}
}
// return C[n][k];
}
// Function to find the number of ways
function numberOfWays(x , y , z) {
// Function to pre-compute
binomialCoeff(Math.max(x, Math.max(y, z)));
// Sum the Zci
var sum = 0;
for (i = 1; i <= z; i++) {
sum = (sum + C[z][i]);
}
// Iterate for second position
var sum1 = 0;
for (i = 1; i <= y; i++) {
// Iterate for first position
for (j = i + 1; j <= x; j++) {
sum1 = (sum1 + (C[y][i] * C[x][j]));
}
}
// Multiply the common Combination value
sum1 = (sum * sum1);
return sum1;
}
// Driver Code
var x = 3;
var y = 2;
var z = 1;
document.write(numberOfWays(x, y, z));
// This code contributed by aashish1995 </script> |
Output:
9
Time Complexity: O(max(x, y, z)^2), due to the computation of binomial coefficients using dynamic programming.
Auxiliary Space: O(max(x, y, z)^2) to store the precomputed binomial coefficients in the 2D array C.