Ways to select one or more pairs from two different sets️‍🔥

Q: What is Ways to select one or more pairs from two different sets?

In this article, we will learn Ways to select one or more pairs from two different sets,This free Data Structures and Algorithms tutorial for complete beginners will help you learn Data Structures and Algorithms from scratch.

Given two positive numbers ‘n’ and ‘m’ (n <= m) which represent total number of items of first and second type of sets respectively. Find total number of ways to select at-least one pair by picking one item from first type(I) and another item from second type(II). In any arrangement, an item should not be common between any two pairs. Note: Since answer can be large, output it in modulo 1000000007.

Input: 2 2
Output: 6
Explanation
Let's denote the items of I type
as a, b and II type as c, d i.e,
Type I -  a, b
Type II - c, d
Ways to arrange one pair at a time
1. a --- c
2. a --- d
3. b --- c
4. b --- d
Ways to arrange two pairs at a time
5. a --- c, b --- d
6. a --- d, b --- c

Input: 2 3
Output: 12

Input: 1 2
Output: 2

The approach is simple, we only need the combination of choosing ‘i‘ items from ‘n‘ type and ‘i‘ items from ‘m‘ type and multiply them(Rule of product) where ‘i‘ varies from 1 to ‘n‘. But we can also permute the resultant product in ‘i’ ways therefore we need to multiply with i!. After that take the sum(Rule of sum) of all resultant product to get the final answer.

[Tex]\implies\displaystyle \sum_{i=1}^{\text{n}} \frac{n!}{i!(n-i)!}\cdot \frac{m!}{i!(m-i)!}\cdot i![/Tex][Tex]\implies\displaystyle\sum_{i=1}^{\text{n}} \frac{{}^n\text P_i\cdot{}^m\text P_i}{i}[/Tex]

C++

Java

Python3

Javascript

C#

// C# program to find total no. of ways
// to form a pair in two different set
using System;
 
class Program {
    static readonly long mod = 1000000007;
    //Iterative Function to calculate (x^y)%p in O(log y)
    static long Power(long x, long y, long p) {
        long res = 1; // Initialize result
        x %= p; // Update x if it is more than or equal to p
 
        // If y is odd, multiply x with result
        while (y > 0) {
            if ((y & 1) == 1) {
                res = (res * x) % p;
            }
             
            // y must be even now
            y >>= 1;
            x = (x * x) % p;
        }
 
        return res;
    }
 
    // Pre-calculate factorial and Inverse of number
    static (long[], long[]) PreCalculate(int n) {
        long[] fact = new long[n + 1];
        long[] inverseMod = new long[n + 1];
 
        fact[0] = inverseMod[0] = 1;
 
        for (int i = 1; i <= n; i++) {
            fact[i] = (fact[i - 1] * i) % mod;
            inverseMod[i] = Power(fact[i], mod - 2, mod);
        }
 
        return (fact, inverseMod);
    }
 
    // utility function to calculate nCr
    static long nPr(int a, int b, long[] fact, long[] inverseMod) {
        return (fact[a] * inverseMod[a - b]) % mod;
    }
 
    static int CountWays(int n, int m) {
        // Pre-calculate factorial and inverse of number
        (long[] fact, long[] inverseMod) = PreCalculate(m);
 
        // Initialize answer
        long ans = 0;
 
        for (int i = 1; i <= n; i++) {
            ans += (((nPr(n, i, fact, inverseMod)
                      * nPr(m, i, fact, inverseMod)) % mod)
                    * inverseMod[i]) % mod;
 
            if (ans >= mod) {
              ans %= mod;
            }
        }
 
        return (int)ans;
    }
    // Derive code 
    static void Main(string[] args) {
        int n = 2;
        int m = 2;
        Console.WriteLine(CountWays(n, m));
    }
}
 
// This code is contributed by Shivhack999

Output:

Time complexity: O(m*log(mod)) Auxiliary space: O(m)