Answer:
The answers are: 1 and 1.
Step-by-step explanation:
I believe they are how combinations are written. In other words:
[tex]_{n}C_{r}=(\frac{n}{r})=n!/(r!(n-r)!)[/tex] (Ignore the fraction bar, I can't really format it properly).
Anyways, simply find the combination of each:
[tex](\frac{9}{0})=9!/(0!(9-0)!)=9!/9!=1[/tex]
[tex](\frac{9}{9} )=9!/(9!(9-9)!)=9!/9!(0!)=9!/9!=1[/tex]
Note that 0! equals 1.
