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

• Difficulty Level : Hard
• Last Updated : 13 Jul, 2022

We need to make a string of size n. Each character of the string is either â€˜Râ€™, â€˜Bâ€™ or â€˜Gâ€™. In the final string there needs to be at least r number of â€˜Râ€™, at least b number of â€˜Bâ€™ and at least g number of â€˜Gâ€™ (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 â€˜Gâ€™ >= 1,
No. of â€˜Bâ€™ >= 1 and
(No. of â€˜Râ€™) + (No. of â€˜Bâ€™) + (No. of â€˜Gâ€™) = n
then following cases are possible:
1. RBGR and its 12 permutation
2. RBGB and its 12 permutation
3. RBGG and its 12 permutation

Recommended Practice

Implementation:

## C++

 `// C++ program to count number of possible strings``// with n characters.``#include``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;``}`

## Java

 `// 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));``    ``}``}`

## Python3

 `# 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`

## C#

 `// 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

 ``

## Javascript

 ``

Output

`22`

Time Complexity: O(n*n).
Auxiliary Space: O(n).

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.

My Personal Notes arrow_drop_up