I am trying to come up with a general formula. First let me give a simple example.
Let there be 3 numbers: a, b, and c.
I multiply any two of them to get:
1. a*b
2. b*c
3. c*a .......[A]
Exactly three ways to multiply any two of them. Right! Next I add any two of above without repetition. I get:
4. a*b + b*c
5. a*b + c*a
6. b*c + c*a
Hence I get 3 combinations. Please observe that I added the a*b to all the multiples in [A] until I exhausted all the combinations. Only then I chose b*c. Next I add all of the results in [A] to get:
7. a*b + b*c + c*a
Now I stop. In total, I get 7 combinations.
Let me try with 4 numbers, say a, b, c, and d.
There are four ways to multiply any three of four numbers. Right! I multiply any three of four numbers to get:
1. a*b*c
2. b*c*d
3. c*d*a
4. d*a*b ......[B]
Next I add any two of above without repetition. I get:
5. a*b*c + b*c*d
6. a*b*c + c*d*a
7. a*b*c + d*a*b
8. b*c*d + c*d*a
9. b*c*d + d*a*b
10. c*d*a + d*a*b
Hence I get 6 combinations. Please observe that I added a*b*c to all the multiples until I exhausted all the combinations. Only then I chose b*c*d and later c*d*a. Next I add any 3 in [B] to get:
11. a*b*c + b*c*d + c*d*a
12. a*b*c + b*c*d + d*a*b
13. a*b*c + c*d*a + d*a*b
14. b*c*d + c*d*a + d*a*b
I get 4 such additions. No more. Right! Lastly I add all four in [B] to get:
15. a*b*c + b*c*d + c*d*a + d*a*b
Now I stop. In total, I get 15 combinations with 4 numbers.
Now I am trying with a set of k numbers.
I can multiply any of (k-1) numbers in the set to get exactly k multiples. Hence k combinations.
Next I choose the first multiple and add it to remaining (k-1) multiples to get (k-1) combinations for addition. I already got all the possible additions with the first multiple. So I strike out the first multiple. Now I chose the second multiple and add it to remaining (k-2) multiples to get (k-2) combinations of addition. Now I strike out the second multiple. I continue with the remaining (k-3) multiples and so on..
I get following number of combinations: (k-1) + (k-2) + (k-3) + ... + (k - (k-1)) = k*(k - 1) - (1 + 2 + 3 + ... + (k-1)) = k*(k - 1) - (Sum of (k-1) natural numbers) = k*(k-1) - (k-1)k/2 = k(k-1)/2
Next I need to choose any 3 multiples non repetitively and find their combinations until I have added all the k multiples.