Given n, r, and K. The task is to find the number of permutations of 
different things taken at a time such that specific things always occur together.
Examples:
Input : n = 8, r = 5, k = 2 Output : 960
Input : n = 6, r = 2, k = 2 Output : 2
Approach:
- A bundle of
specific things can be put in r places in (r – k + 1) ways . - k specific things in the bundle can be arranged themselves into k! ways.
- Now (n – k) things will be arranged in (r – k) places in
ways.
Thus, using the fundamental principle of counting, the required number of permutations will be:
Below is the implementation of the above approach:
C++
// CPP program to find the number of permutations of// n different things taken r at a time// with k things grouped together#include <bits/stdc++.h>using namespace std;// Function to find factorial// of a numberint factorial(int n){ int fact = 1; for (int i = 2; i <= n; i++) fact = fact * i; return fact;}// Function to calculate p(n, r)int npr(int n, int r){ int pnr = factorial(n) / factorial(n - r); return pnr;}// Function to find the number of permutations of// n different things taken r at a time// with k things grouped togetherint countPermutations(int n, int r, int k){ return factorial(k) * (r - k + 1) * npr(n - k, r - k);}// Driver codeint main(){ int n = 8; int r = 5; int k = 2; cout << countPermutations(n, r, k); return 0;} |
Java
// Java program to find the number of permutations of// n different things taken r at a time// with k things grouped togetherclass GFG{// Function to find factorial// of a numberstatic int factorial(int n){ int fact = 1; for (int i = 2; i <= n; i++) fact = fact * i; return fact;}// Function to calculate p(n, r)static int npr(int n, int r){ int pnr = factorial(n) / factorial(n - r); return pnr;}// Function to find the number of permutations of// n different things taken r at a time// with k things grouped togetherstatic int countPermutations(int n, int r, int k){ return factorial(k) * (r - k + 1) * npr(n - k, r - k);}// Driver codepublic static void main(String[] args){ int n = 8; int r = 5; int k = 2; System.out.println(countPermutations(n, r, k));}}// this code is contributed by mits |
Python3
# Python3 program to find the number of permutations of # n different things taken r at a time # with k things grouped together # def to find factorial # of a number def factorial(n): fact = 1; for i in range(2,n+1): fact = fact * i; return fact; # def to calculate p(n, r) def npr(n, r): pnr = factorial(n) / factorial(n - r); return pnr; # def to find the number of permutations of # n different things taken r at a time # with k things grouped together def countPermutations(n, r, k): return int(factorial(k) * (r - k + 1) * npr(n - k, r - k)); # Driver code n = 8; r = 5; k = 2; print(countPermutations(n, r, k)); # this code is contributed by mits |
C#
// C# program to find the number of// permutations of n different things// taken r at a time with k things // grouped together using System;class GFG{ // Function to find factorial // of a number static int factorial(int n) { int fact = 1; for (int i = 2; i <= n; i++) fact = fact * i; return fact; } // Function to calculate p(n, r) static int npr(int n, int r) { int pnr = factorial(n) / factorial(n - r); return pnr; } // Function to find the number of // permutations of n different // things taken r at a time with // k things grouped together static int countPermutations(int n, int r, int k) { return factorial(k) * (r - k + 1) * npr(n - k, r - k); } // Driver code static void Main() { int n = 8; int r = 5; int k = 2; Console.WriteLine(countPermutations(n, r, k)); } } // This code is contributed by mits |
PHP
<?php// PHP program to find the number // of permutations of n different // things taken r at a time// with k things grouped together// Function to find factorial// of a numberfunction factorial($n){ $fact = 1; for ($i = 2; $i <= $n; $i++) $fact = $fact * $i; return $fact;}// Function to calculate p(n, r)function npr($n, $r){ $pnr = factorial($n) / factorial($n - $r); return $pnr;}// Function to find the number of // permutations of n different // things taken r at a time// with k things grouped togetherfunction countPermutations($n, $r, $k){ return factorial($k) * ($r - $k + 1) * npr($n - $k, $r - $k);}// Driver code$n = 8;$r = 5;$k = 2;echo countPermutations($n, $r, $k);// This code is contributed by mits?> |
Javascript
<script>// JavaScript program to find the number of permutations of // n different things taken r at a time // with k things grouped together // Function to find factorial // of a number function factorial(n) { let fact = 1; for (let i = 2; i <= n; i++) fact = fact * i; return fact; } // Function to calculate p(n, r) function npr(n, r) { let pnr = Math.floor(factorial(n) / factorial(n - r)); return pnr; } // Function to find the number of permutations of // n different things taken r at a time // with k things grouped together function countPermutations(n, r, k) { return factorial(k) * (r - k + 1) * npr(n - k, r - k); } // Driver code let n = 8; let r = 5; let k = 2; document.write(countPermutations(n, r, k)); // This code is contributed by Surbhi Tyagi.</script> |
Output:
960
Time Complexity: O(n)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
… [Trackback]
[…] Info to that Topic: geeksforgeeks.org/permutations-of-n-things-taken-r-at-a-time-with-k-things-together/ […]