Question

help.

Answer 1

number 8.

Answer 2

sum of 11 6+5
sum(s) of 7 6+1 , 5+2, 4+3

the rest is
sum of 2 1+1
sum of 3 1+2
sum(s) of 4 1+3, 2+2
sum(s) of 5 1+4, 2+3
sum(s) of 6 1+5, 2+4, 3+3 (skipping the sum of 7)
sum(s) of 8 2+6, 3+5, 4+4
sum(s) of 9 3+6, 4+5
sum(s) of 10 4+6, 5+5
sum of 12 6+6

so all the sums possible including 11 and 7, are
21 sums

and there are 4 sums that can be 11 and 7,

So 4/21 (I am not 100% sure on this one)

Answer 3

unless this is nPr, and you would count twice 5+6 and 6+5... agh idk...