Recursion function -counting permuation and ignoring permutation - c++

Question

I am given this problem:
We are going over recursion in my class and I do not quite understand it, I was wondering if someone can help me with this problem
let c(n) be the number of different group integers that can be chosen from the integers 1 through n-1, so that the integers in each group add up to n (for example, n=4=[1+1+1+1]=[1+1+2]=[2+2]). Write a recursive definition for c(n) under the following variations:
a) You count permutations. For example, 1,2,1 and 1,1,2 are two groups that each add up to 4
b)you ignore permutations
I know permutations is how many ways you can arrange a set of numbers, so is my code below correct? I get an answer of 7?
Here is my code for part a:
int recurse (int n);
int main(){
int a=4;
int sum_perm;
sum_perm=recurse(a);
cout<<sum_perm-1<<endl;
//Can I do -1 here because it should be from a group of integers from 1 to n-1?
return 0;
}
int recurse(int n)
{
int sum = 1;
if (n == 1){
return 1;
}
for(int i = 1; i < n; i++){
sum += recurse(n - i);
}
return sum;
}
For part B, if I am not counting permutations, what am I counting?
Here is my code for part b:
int without (int n, int max);
int main(){
int a=4, b =3;
int sum_without;
sum_without=without(a,b);
cout<<sum_without<<endl;
system("Pause");
return 0;
}
int without(int n, int max)
{
if(n == 1 || max == 1){
return 1;
}
else if (n == max){
return 1 + without(n, n-1);
}
else{
return without(n,max-1) + without(n-max, max);
}
}

Answer 1

You don't show any code to generate the combinations of numbers that produce a sum. Link to wiki article about partitions .
In this case, the goal is to count the number of combinations and/or permutations, which might be possible without actually generating a set of combinations. Not sure if recursion helps here, but you can convert any loop into recursion if you pass enough variables.
Example "partitions"
1 combination that sums to 1:
1
2 combinations that sum to 2:
1 1
2
3 combinations that sum to 3:
1 1 1
1 2
3
5 combinations that sum to 4:
1 1 1 1
1 1 2
1 3
2 2
4
7 combinations that sum to 5:
1 1 1 1 1
1 1 1 2
1 1 3
1 2 2
1 4
2 3
5
11 combinations of numbers that sum to 6:
1 1 1 1 1 1
1 1 1 1 2
1 1 1 3
1 1 2 2
1 1 4
1 2 3
2 2 2
1 5
2 4
3 3
6

Answer 2

I would recommend combinations directly being considered. While it seems like the more difficult case a simple rule makes it trivial.
Numbers calculated are in decreasing order
This requires you to track the last number, but ensures you don't calculate 1 5 and then 5 1, as the former is impossible.