Rivits
The number of combinations of 2 from 10 choices is 45 - Still a bit of a job though
I Know what has happened the perm 0000010001 for example has been counted twice.
So we produce a work table as follows:
1100000000 0110000000 0011000000 0001100000 0000110000 .........
1010000000 0101000000 0010100000 0001010000 0000101000
1001000000 0100100000 0010010000 0001001000 0000100100
1000100000 0100010000 0010001000 0001000100 0000100010
1000010000 0100001000 0010000100 0001000010 0000100001
1000001000 0100000100 0010000010 0001000001
1000000100 0100000010 0010000001
1000000010 0100000001
1000000001
You can see how its going to turn out but you will get there. 9+8+7+6+5+4+3+2+1 i.e 45
You can do this for both switch sets but the formatting is too difficult to post.
CAT III
Edit: Unixman I know what you were thinking of but its not a single digit binary problem.
Last edited by Guest 112233; 26th August 2013 at 08:42.