sorry for some of the confusion, what i meant in a few of the above equations was

(aCm) plus (bCm)
---------------- = P(x)
(a plus b)Cm

,

P(x)=

(5C2) plus (5C2)
---------------- =0.444
10C2

also, i definately agree with alexb, but i still think it can be done in a much simpler way without having to go though all the possibilities of the different combinations of odd and even numbers to get an even sum.
Ben, i can also see how it might be easier to find the probability of odd sums and subtract from the total probability, however, the question still remains of how to go about this without going through every possible combination of how you can get an odd sum. it can be easily done with my example but u must also take into consideration of how to do it if "n" is a much greater # like 399,752 fo example. cant exactly count all the different combinations for that now can ya?!?! this is where most of my problems now lie!

i've also made another small example where multiplication is used rather than addition btw even and odd sums...

1,2,3,4,5,6,7,8,9,10
-n=10
choose 5
-m=5

P(x)=
(5C5)(5C0) PLUS (5C3)(5C2) PLUS (5C1)(5C4)
------------------------------------------
10C5

= some # i cant remember, whatever

this is how i've found the total ways of being able to get an even sum but for larger #s, there just has to be another way to figure it out!
can anyone help me out here???