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.
C++
// C++ program to find total no. of ways // to form a pair in two different set #include <bits/stdc++.h> using namespace std; // initialize global variable so that // it can access by preCalculate() and // nCr() function int* fact, *inverseMod; const int mod = 1e9 + 7; /* Iterative Function to calculate (x^y)%p in O(log y) */int power(int x, int y, int p) { int res = 1; // Initialize result x = x % p; // Update x if it is more than or // equal to p while (y) { // If y is odd, multiply x with result if (y & 1) res = (1LL * res * x) % p; // y must be even now y = y >> 1; // y = y/2 x = (1LL * x * x) % p; } // trace(res); return res; } // Pre-calculate factorial and // Inverse of number void preCalculate(int n) { fact[0] = inverseMod[0] = 1; for (int i = 1; i <= n; ++i) { fact[i] = (1LL * fact[i - 1] * i) % mod; inverseMod[i] = power(fact[i], mod - 2, mod); } } // utility function to calculate nCr int nPr(int a, int b) { return (1LL * fact[a] * inverseMod[a - b]) % mod; } int countWays(int n, int m) { fact = new int[m + 1]; inverseMod = new int[m + 1]; // Pre-calculate factorial and // inverse of number preCalculate(m); // Initialize answer int ans = 0; for (int i = 1; i <= n; ++i) { ans += (1LL * ((1LL * nPr(n, i) * nPr(m, i)) % mod) * inverseMod[i]) % mod; if (ans >= mod) ans %= mod; } return ans; } // Driver program int main() { int n = 2, m = 2; cout << countWays(n, m); return 0; } |
Java
// Java program to find total // no. of ways to form a pair // in two different set public class Test { static long[] fact; static long[] inverseMod; static int mod=1000000007; /* Iterative Function to calculate (x^y)%p in O(log y) */ static long power(long x, int y, int p) { // Initialize result long res = 1; // Update x if it is more than or // equal to p x = x % p; while (y!=0) { // If y is odd, multiply // x with result if ((y & 1)!=0) res = (1 * res * x) % p; } // y must be even now y = y >> 1; x = (1 * x * x) % p; } return res; } // Pre-calculate factorial and // Inverse of number public static void preCalculate(int n) { //int fact[]=new long[n]; // int inverseMod[]=new long[n]; fact[0] = 1; inverseMod[0] = 1; for (int i = 1; i <= n; i++) { fact[i] = (1 * fact[i - 1] * i) % mod; inverseMod[i] = power(fact[i], mod - 2, mod); } } // utility function to calculate nCr public static long nPr(int a, int b) { return (1 * fact[a] * inverseMod[a - b]) % (long)mod; } public static int countWays(int n, int m) { fact = new long[m + 1]; inverseMod = new long[m + 1]; // Pre-calculate factorial and // inverse of number preCalculate(m); // Initialize answer long ans = 0; for (int i = 1; i <= n; i++) { ans = ans+(1 * ((1 * nPr(n, i)* nPr(m, i)) % mod)* (inverseMod[i])) % mod; if (ans >= mod) ans %= mod; } return (int)ans; } /* Driver program */ public static void main(String[] args) { int n = 2, m = 2; System.out.println(countWays(n, m)); } } // This code is contributed by Gitanjali |
Output:
6
Time complexity: O(m*log(mod))
Auxiliary space: O(m)
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