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

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

  1. As R, B and G have to be included atleast for given no. of times. Remaining values = n -(r + b + g).
  2. Make all combinations for the remaining values.
  3. Consider each element one by one for the remaining values and sum up all the permuations.
  4. Return total no. of permutations of all the combinations.

C++

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// C++ program to count number of possible strings
// with n characters.
#include<bits/stdc++.h>
using namespace std;
  
// Function to calculate number of strings
int 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 code
int main()
{
    int n = 4, r = 2;
    int b = 0, g = 1;
    cout << possibleStrings(n, r, b, g);
    return 0;
}

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Java

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// 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));
    }
}

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Python3

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# Python 3 program to count number of 
# possible strings with n characters.
  
# Function to calculate number of strings
def 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 code
if __name__ == '__main__':
    n = 4
    r = 2
    b = 0
    g = 1
    print(int(possibleStrings(n, r, b, g)))
      
# This code is contributed by
# Sanjit_Prasad

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C#

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// 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.

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PHP

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<?php
// PHP 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
    $fact[0] = 1;
    for ($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.
?>

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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 contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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