Hi XOR LX ,
MATCH(31,MMULT(N({1;2;3;4;5;6;7;8;9;10;11}>={1,2,3,4,5,6,7,8,9,10,11}),B2:B12),1)
Gives a result, some numbers should at least, to get a total of 31.
Second,
On your site, as there are numbers from 1 to 10, how many possible combinations to the amount of 20, the result is 8.
The question is it possible that any combination will result in one cell, that is, (5,7,8) = CB1 etc,
If you take ten numbers, each made up of a combination of three numbers, we come to 720 options,=PERMUT(10,3) or,10*9*8,
With three columns,
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
8 8 8
9 9 9
10 10 10
Included in the calculation, the unique combination, like 1,2,3 or 5,8,6
Regards
David