Count of all possible pairs of disjoint subsets of integers from 1 to N

Given an integer N. Consider the set of first N natural numbers A = {1, 2, 3, …, N}. Let M and P be two non-empty subsets of A. The task is to count the number of unordered pairs of (M, P) such that M and P are disjoint sets. Note that the order of M and P doesn’t matter.

Examples:

Input: N = 3
Output: 6
The unordered pairs are ({1}, {2}), ({1}, {3}),
({2}, {3}), ({1}, {2, 3}), ({2}, {1, 3}), ({3}, {1, 2}).

Input: N = 2
Output: 1

Input: N = 10
Output: 28501

Approach:

1. Lets assume there are only 6 elements in the set {1, 2, 3, 4, 5, 6}.
2. When you count the number of subsets with 1 as one of the element of first subset, it comes out to be 211.
3. Counting number of subsets with 2 being one of the element of first subset, it comes out to be 65, because 1’s not included as order of sets doesn’t matter.
4. Counting number of subset with 3 being one of the element of first set it comes out to be 65, here a pattern can be observed.
5. Pattern:
   

   5 = 3 * 1 + 2
   19 = 3 * 5 + 4
   65 = 3 * 19 + 8
   211 = 3 * 65 + 16
   S(n) = 3 * S(n-1) + 2^{(n – 2)}

6. Expanding it until n->2 (means numbers of elements n-2+1=n-1)
   2^(n-2) * 3⁽⁰⁾ + 2^{(n – 3)} * 3¹ + 2^{(n – 4)} * 3² + 2^{(n – 5)} * 3³ + … + 2⁽⁰⁾ * 3^{(n – 2)}
   From Geometric progression, a + a * r⁰ + a * r¹ + … + a * r^{(n – 1)} = a * (rⁿ – 1) / (r – 1)
7. S(n) = 3^{(n – 1)} – 2^{(n – 1)}. Remember S(n) is number of subsets with 1 as a one of the elements of first subset but to get the required result, Denoted by T(n) = S(1) + S(2) + S(3) + … +S(n)
8. It also forms a Geometric progression, so we calculate it by formula of sum of GP
   T(n) = (3ⁿ – 2^{n + 1} + 1)/2
9. As we require T(n) % p where p = 10⁹ + 7
   We have to use Fermats’s little theorem
   a^-1 = a^{(m – 2)} (mod m) for modular division

   Below is the implementation of the above approach:
   

   C++

   
   

   
   
   

   // C++ implementation of the approach    #include <bits/stdc++.h> using namespace std; #define p 1000000007    // Modulo exponentiation function long long power(long long x, long long y) {     // Function to calculate (x^y)%p in O(log(y))     long long res = 1;     x = x % p;        while (y > 0) {         if (y & 1)             res = (res * x) % p;         y = y >> 1;         x = (x * x) % p;     }        return res % p; }    // Driver function int main() {     long long n = 3;        // Evaluating ((3^n-2^(n+1)+1)/2)%p     long long x = (power(3, n) % p + 1) % p;        x = (x - power(2, n + 1) + p) % p;        // From  Fermats’s little theorem     // a^-1 ? a^(m-2) (mod m)        x = (x * power(2, p - 2)) % p;     cout << x << "\n"; }

   Java

   
   

   
   
   

   // Java implementation of the approach class GFG  {  static int p = 1000000007;    // Modulo exponentiation function static long power(long x, long y) {     // Function to calculate (x^y)%p in O(log(y))     long res = 1;     x = x % p;        while (y > 0)     {         if (y % 2 == 1)             res = (res * x) % p;         y = y >> 1;         x = (x * x) % p;     }     return res % p; }    // Driver Code public static void main(String[] args)  {     long n = 3;        // Evaluating ((3^n-2^(n+1)+1)/2)%p     long x = (power(3, n) % p + 1) % p;        x = (x - power(2, n + 1) + p) % p;        // From Fermats's little theorem     // a^-1 ? a^(m-2) (mod m)        x = (x * power(2, p - 2)) % p;     System.out.println(x); } }    // This code is contributed by Rajput-Ji

   Python3

   
   

   
   
   

   # Python3 implementation of the approach p = 1000000007    # Modulo exponentiation function def power(x, y):            # Function to calculate (x^y)%p in O(log(y))     res = 1     x = x % p        while (y > 0):         if (y & 1):             res = (res * x) % p;         y = y >> 1         x = (x * x) % p        return res % p    # Driver Code n = 3    # Evaluating ((3^n-2^(n+1)+1)/2)%p x = (power(3, n) % p + 1) % p    x = (x - power(2, n + 1) + p) % p    # From Fermats’s little theorem # a^-1 ? a^(m-2) (mod m) x = (x * power(2, p - 2)) % p    print(x)    # This code is contributed by Mohit Kumar

   C#

   
   

   
   
   

   // C# implementation of the approach using System;    class GFG { static int p = 1000000007;    // Modulo exponentiation function static long power(long x, long y) {     // Function to calculate (x^y)%p in O(log(y))     long res = 1;     x = x % p;        while (y > 0)     {         if (y % 2 == 1)             res = (res * x) % p;         y = y >> 1;         x = (x * x) % p;     }     return res % p; }    // Driver Code static public void Main () {     long n = 3;            // Evaluating ((3^n-2^(n+1)+1)/2)%p     long x = (power(3, n) % p + 1) % p;            x = (x - power(2, n + 1) + p) % p;            // From Fermats's little theorem     // a^-1 ? a^(m-2) (mod m)            x = (x * power(2, p - 2)) % p;     Console.Write(x); } }    // This code is contributed by ajit.

   Output:

   6
   

   

   
   
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