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Given three numbers R, G and B and letters ‘R’, ‘G’ and ‘B’ only. The goal is to find the count of possible strings that can be made using at least R Rs, at least G Gs and at least B Bs in it. The numbers R, G and B have sum less than or equal to the length of strings possible.

**For Example**

R = 1, G = 1, B = 1 length=3

Count of number of strings (made of R, G and B) using given combination are − 6

The possible strings will be : “RGB”, “RBG”, “BRG”, “BGR”, “GRB”, “GBR”. That is permutations of RGB.

R = 2, G = 0, B = 2 length=4

Count of number of strings (made of R, G and B) using given combination are − 6

The possible strings will be : “RRBB”, “BBRR”, “RBRB”, “BRBR”, “RBBR”, “BRRB”.

**Approach used in the below program is as follows** −

In this approach we will first take letters for R, B and G number of times. Now check for remaining length − (R+G+B) letters make all permutations and combinations and add to count.

Take integer values R, G, B as input.

Take size as length of the strings to be made.

Function combination(int R, int G, int B, int size) takes all inputs and returns the count of number of strings (made of R, G and B) using a given combination.

Take the initial count as 0.

Take remaining count as temp=size − (R + G + B).

Take an array arr of length size+1 to take values of permutations.

Initially store factorials of i in arr[i]. Using for loop from i=0 to i=size. Set arr[i]=arr[i−1]*i.

For calculation of combinations traverse arr[] again using two for loops.

For i=0 to temp and j=0 to temp−i, compute temp_2 = temp − (i + j).

Take temp_3 = arr[i + R] * arr[j + B] * arr[temp_2 + G].

Now add arr[size] / temp_3 to count.

At the end of all for loops we will have counted as the required number of strings possible.

Return count as result.

#include<bits/stdc++.h> using namespace std; int combination(int R, int G, int B, int size){ int count = 0; int temp = size − (R + G + B); int arr[size+1]; arr[0] = 1; for (int i = 1; i <= size; i++){ arr[i] = arr[i − 1] * i; } for (int i = 0; i <= temp; i++){ for (int j = 0; j <= temp−i; j++){ int temp_2 = temp − (i + j); int temp_3 = arr[i + R] * arr[j + B] * arr[temp_2 + G]; count += arr[size] / temp_3; } } return count; } int main(){ int R = 2, G = 1, B = 1; int size = 4; cout<<"Count of number of strings (made of R, G and B) using given combination are: "<<combination(R, G, B, size); return 0; }

If we run the above code it will generate the following output −

Count of number of strings (made of R, G and B) using given combination are: 12

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