Count number of strings (made of R, G and B) using given combination - GeeksforGeeks

08 Jun, 2021
We need to make a string of size n. Each character of the string is either ‘R’, ‘B’ or ‘G’. In the final string there needs to be at least r number of ‘R’, at least b number of ‘B’ and at least g number of ‘G’ (such that r + g + b <= n). We need to find number of such strings possible.Examples:  
Input : n = 4, r = 1, 
        b = 1, g = 1.
Output: 36 
No. of 'R' >= 1, 
No. of ‘G’ >= 1, 
No. of ‘B’ >= 1 and 
(No. of ‘R’) + (No. of ‘B’) + (No. of ‘G’) = n
then following cases are possible:
1. RBGR and its 12 permutation
2. RBGB and its 12 permutation
3. RBGG and its 12 permutation
Hence answer is 36.
Asked in : Directi 
 
As R, B and G have to be included atleast for given no. of times. Remaining values = n -(r + b + g).Make all combinations for the remaining values.Consider each element one by one for the remaining values and sum up all the permutations.Return total no. of permutations of all the combinations.
As R, B and G have to be included atleast for given no. of times. Remaining values = n -(r + b + g).
Make all combinations for the remaining values.
Consider each element one by one for the remaining values and sum up all the permutations.
Return total no. of permutations of all the combinations.
 
C++
Java
Python3
C#
PHP
Javascript
// C++ program to count number of possible strings// with n characters.#include<bits/stdc++.h>using namespace std; // Function to calculate number of stringsint possibleStrings( int n, int r, int b, int g){    // Store factorial of numbers up to n    // for further computation    int fact[n+1];    fact[0] = 1;    for (int i = 1; i <= n; i++)        fact[i] = fact[i-1] * i;     // Find the remaining values to be added    int left = n - (r+g+b);    int sum = 0;     // Make all possible combinations    // of R, B and G for the remaining value    for (int i = 0; i <= left; i++)    {        for (int j = 0; j<= left-i; j++)        {            int k = left - (i+j);             // Compute permutation of each combination            // one by one and add them.            sum = sum + fact[n] /                       (fact[i+r]*fact[j+b]*fact[k+g]);        }    }     // Return total no. of strings/permutation    return sum;} // Drivers codeint main(){    int n = 4, r = 2;    int b = 0, g = 1;    cout << possibleStrings(n, r, b, g);    return 0;}
// Java program to count number of possible// strings with n characters. class GFG{         //Function to calculate number of strings    static int possibleStrings( int n, int r, int b, int g)    {     // Store factorial of numbers up to n     // for further computation     int fact[] = new int[n+1];     fact[0] = 1;     for (int i = 1; i <= n; i++)         fact[i] = fact[i-1] * i;      // Find the remaining values to be added     int left = n - (r+g+b);     int sum = 0;      // Make all possible combinations     // of R, B and G for the remaining value     for (int i = 0; i <= left; i++)     {         for (int j = 0; j<= left-i; j++)         {             int k = left - (i+j);              // Compute permutation of each combination             // one by one and add them.             sum = sum + fact[n] /                        (fact[i+r]*fact[j+b]*fact[k+g]);         }     }      // Return total no. of strings/permutation     return sum;    }     //Drivers code    public static void main(String []args)    {        int n = 4, r = 2;         int b = 0, g = 1;         System.out.println(possibleStrings(n, r, b, g));    }}
# Python 3 program to count number of# possible strings with n characters. # Function to calculate number of stringsdef possibleStrings(n, r, b, g):         # Store factorial of numbers up to n    # for further computation    fact = [0 for i in range(n + 1)]    fact[0] = 1    for i in range(1, n + 1, 1):        fact[i] = fact[i - 1] * i     # Find the remaining values to be added    left = n - (r + g + b)    sum = 0     # Make all possible combinations of    # R, B and G for the remaining value    for i in range(0, left + 1, 1):        for j in range(0, left - i + 1, 1):            k = left - (i + j)             # Compute permutation of each            # combination one by one and add them.            sum = (sum + fact[n] / (fact[i + r] *                         fact[j + b] * fact[k + g]))         # Return total no. of    # strings/permutation    return sum # Driver codeif __name__ == '__main__':    n = 4    r = 2    b = 0    g = 1    print(int(possibleStrings(n, r, b, g)))     # This code is contributed by# Sanjit_Prasad
// C# program to count number of possible// strings with n characters.using System; class GFG{     //Function to calculate number of strings    static int possibleStrings( int n, int r,                                int b, int g)    {        // Store factorial of numbers up to n        // for further computation        int[] fact = new int[n + 1];        fact[0] = 1;             for (int i = 1; i <= n; i++)            fact[i] = fact[i - 1] * i;         // Find the remaining values to be added        int left = n - (r + g + b);        int sum = 0;         // Make all possible combinations        // of R, B and G for the remaining value        for (int i = 0; i <= left; i++)        {            for (int j = 0; j <= left - i; j++)            {                int k = left - (i + j);                 // Compute permutation of each combination                // one by one and add them.                sum = sum + fact[n] / (fact[i + r] *                        fact[j + b] * fact[k + g]);            }        }         // Return total no. of strings/permutation        return sum;    }     //Drivers code    public static void Main()    {        int n = 4, r = 2;        int b = 0, g = 1;        Console.WriteLine(possibleStrings(n, r, b, g));    }} // This Code is contributed by Code_Mech.
<?php// PHP program to count number// of possible strings with// n characters. // Function to calculate// number of stringsfunction possibleStrings( $n, $r, $b, $g){         // Store factorial of    // numbers up to n for    // further computation    $fact[0] = 1;    for ($i = 1; $i <= $n; $i++)        $fact[$i] = $fact[$i - 1] * $i;     // Find the remaining    // values to be added    $left = $n - ($r + $g + $b);    $sum = 0;     // Make all possible combinations    // of R, B and G for the remaining value    for ($i = 0; $i <= $left; $i++)    {        for ($j = 0; $j <= $left - $i; $j++)        {            $k = $left - ($i+$j);             // Compute permutation of            // each combination one            // by one and add them.            $sum = $sum + $fact[$n] /                   ($fact[$i + $r] *                   $fact[$j + $b] *                   $fact[$k + $g]);        }    }     // Return total no. of    // strings/permutation    return $sum;}     // Driver Code    $n = 4; $r = 2;    $b = 0; $g = 1;         echo possibleStrings($n, $r, $b, $g); // This code is contributed by jit_t.?>
<script>// Javascript program to count number of possible// strings with n characters.         // Function to calculate number of strings    function possibleStrings(n,r,b,g)    {        // Store factorial of numbers up to n     // for further computation     let fact = new Array(n+1);     fact[0] = 1;     for (let i = 1; i <= n; i++)         fact[i] = fact[i-1] * i;        // Find the remaining values to be added     let left = n - (r+g+b);     let sum = 0;        // Make all possible combinations     // of R, B and G for the remaining value     for (let i = 0; i <= left; i++)     {         for (let j = 0; j<= left-i; j++)         {             let k = left - (i+j);                // Compute permutation of each combination             // one by one and add them.             sum = sum + fact[n] /                        (fact[i+r]*fact[j+b]*fact[k+g]);         }     }        // Return total no. of strings/permutation     return sum;    }         // Drivers code    let n = 4, r = 2;    let b = 0, g = 1;    document.write(possibleStrings(n, r, b, g));         // This code is contributed by avanitrachhadiya2155     </script>
Output:  
22
To handle n with large numbers, we can use the concept of Large Factorial.This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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Combinatorial
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