String alphabets are only {a, b, c}

Length of string is n. (n>0)

Let's consider what can be the possible cases

1. String is only built with 'a', i.e., n 'a' forms the string.
   Count of such string is: 1
2. String built with one 'b' & n-1 'a'
   Count of such string is: (n/1)=n
   One 'b' can be placed at any of n positions, that's why n number of such strings
3. String built with one 'b', one 'c' and (n-2) 'a'
   Count of such string (n/2)*2=n*(n-1)
   One 'b' and one 'c' can take any of two places out of n and any of 'b' & 'c' can comes first.
4. String built with one 'b', two 'c' and (n-3) 'a'
   Count of such string (n/3)*3=n*(n-1)*(n-2)/2
   One 'b' and two 'c' can take any of three places out of n and there are 3 combinations possible between one 'b' & two 'c'.
5. String built with two 'c' and (n-2) 'a'
   Count of such string (n/2)=n*(n-1)/2
   Two 'c' can take any two of n places.
6. String built with one 'c' and (n-1) 'a'
   Count of such string (n/1)=n
   One 'c' can take any of one places out of n.

Example with explanation

    Let n=2
    Case 1: String is only built with 'a', i.e., n 'a' forms the string
    "aa"
    Total under this category: 1

    Case 2: String built with one 'b' & n-1 'a'
    "ab"
    "ba"
    Total under this category: 2//(n)
    
    Case 3: String built with one 'b', one 'c' and (n-2) 'a'
    "bc"
    "cb"
    Total under this category: 2//(n*(n-1))

    Case 4: String built with one 'b', two 'c' and (n-3) 'a'
    No string in this category
    Total under this category: 0

    Case 5: String built with two 'c' and (n-2) 'a'
    "cc"
    Total under this category: 1//(n*(n-1)/2)

    Case 6: String built with one 'c' and (n-1) 'a'
    "ac"
    "ca"
    Total under this category: 2//(n)

    Total no of strings possible: 1+2+2+0+1+2=8

C++ implementation

#include <bits/stdc++.h>
using namespace std;

int find(int n){
	//total no of string possible(for details check solution part)
	return 1+(n)+n*(n-1)+n*(n-1)*(n-2)/2+n*(n-1)/2+n;
}

int main()
{
	int n;

	cout<<"enter length of string\n";
	cin>>n;
	
	cout<<"Number of string possible under ";
	cout<<"constraints is : "<<find(n)<<endl;

	return 0;
}

Output

First run:
enter length of string
2
Number of string possible under constraints is : 8

Second run:
enter length of string
4
Number of string possible under constraints is : 39