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 
. Hence the following steps can be followed to solve the above problem.
- 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.
* 
.
+ 
+ … 
since that was the common part in both.can be pre-computed using Dynamic Programming to reduce time complexity. The method is discussed here.
Below is the implementation of the above approach.
C++
// 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 DPvoid 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 waysint 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 Codeint main(){ int x = 3; int y = 2; int z = 1; cout << numberOfWays(x, y, z); return 0;} |
Java
// 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
# Python3 program to find the number of# ways to form the group of peoplsC = [[0 for i in range(1000)] for i in range(1000)]# Function to pre-compute the# Combination using DPdef 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 waysdef 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 Codex = 3y = 2z = 1print(numberOfWays(x, y, z))# This code is contributed by Mohit Kumar |
C#
// 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 |
Javascript
<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.
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