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Count number of strings (made of R, G and B) using given combination
• Difficulty Level : Hard
• Last Updated : 28 Jan, 2019

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
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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.

## 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

 ` `

Output:

```22
```

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

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