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

• Difficulty Level : Hard
• Last Updated : 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

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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 permutations.

## 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 = 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 = 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

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## Javascript

 ``

Output:

`22`

To handle n with large numbers, we can use the concept of Large Factorial.
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