Given an integer N where 4 ? N ? 100. There are N lines vertically and N lines horizontally. So, There are N2 intersections. The task is find the number of ways to place 4 items in these N2 positions such that each row and column contain not more than one item.
Examples:
Input: N = 4
Output: 24
Input: N = 5
Output: 600
Approach: The number of ways to choose 4 horizontal lines that will have items from n is nC4. There are n ways to place an item on the first of these lines. Given the place of the first item, there are n – 1 ways to place an item on the second of these lines because one of the vertical lines is already taken. Given the places of the first and second items, there are n – 2 ways to place an item on the third line and the same way n – 3 for the fourth item. The total number of ways to place items on selected 4 horizontal paths is n * (n – 1) * (n – 2) * (n – 3). So the result is nC4 * n * (n – 1) * (n – 2) * (n – 3).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;// Function to return the number of ways// to place 4 items in n^2 positionslong long NumberofWays(int n){ long long x = (1LL * (n) * (n - 1) * (n - 2) * (n - 3)) / (4 * 3 * 2 * 1); long long y = (1LL * (n) * (n - 1) * (n - 2) * (n - 3)); return (1LL * x * y);}// Driver codeint main(){ int n = 4; cout << NumberofWays(n); return 0;} |
Java
// Java implementation of the approachclass GFG{// Function to return the number of ways // to place 4 items in n^2 positions static long NumberofWays(int n) { long x = (1l * (n) * (n - 1) * (n - 2) * (n - 3)) / (4 * 3 * 2 * 1); long y = (1l * (n) * (n - 1) * (n - 2) * (n - 3)); return (1l * x * y); } // Driver code public static void main(String args[]){ int n = 4; System.out.println( NumberofWays(n)); } }// This code is contributed by Arnab Kundu |
Python3
# python implementation of the approach # Function to return the number of ways # to place 4 items in n^2 positions def NumbersofWays(n): x = (n * (n - 1) * (n - 2) * (n - 3)) // (4 * 3 * 2 * 1) y = n * (n - 1) * (n - 2) * (n - 3) return x * y# Driver coden = 4print(NumbersofWays(n))# This code is contributed by Shrikant13 |
C#
// C# implementation of the approach using System;class GFG{// Function to return the number of ways // to place 4 items in n^2 positions public static long NumberofWays(int n){ long x = (1l * (n) * (n - 1) * (n - 2) * (n - 3)) / (4 * 3 * 2 * 1); long y = (1l * (n) * (n - 1) * (n - 2) * (n - 3)); return (1l * x * y);}// Driver code public static void Main(string[] args){ int n = 4; Console.WriteLine(NumberofWays(n));}}// This code is contributed by Shrikant13 |
PHP
<?php// PHP implementation of the approach// Function to return the number of ways// to place 4 items in n^2 positionsfunction NumberofWays($n){ $x = (1 * ($n) * ($n - 1) * ($n - 2) * ($n - 3)) / (4 * 3 * 2 * 1); $y = (1 * ($n) * ($n - 1) * ($n - 2) * ($n - 3)); return (1 * $x * $y);}// Driver code$n = 4;echo NumberofWays($n);// This code is contributed by mits?> |
Javascript
<script> // JavaScript implementation of the approach // Function to return the number of ways // to place 4 items in n^2 positions function NumberofWays(n) { var x = (1 * n * (n - 1) * (n - 2) * (n - 3)) / (4 * 3 * 2 * 1); var y = 1 * n * (n - 1) * (n - 2) * (n - 3); return 1 * x * y; } // Driver code var n = 4; document.write(NumberofWays(n)); </script> |
24
Time Complexity: O(1)
Auxiliary Space: O(1)
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