Q: what is the probability of selecting "m" intergers from (1,2,3,...n)and having the sum of the "m" intergers be even?
(Basically, you're trying to figure out how many even sums you can get when chosing m amount of numbers from 1-n...but in a probability answer)

information:
-(1,2,3,...n)<-- the numbers here start at 1 and are consecutive all the way to "n"
-i would assume that "m" is a positive # and equal to or less then "n"
-an even sum can only be obtained through either the addition of all even sums, an even number of odd sums or a combination of the two (an even number of odd sums and 1+ even numbers)
-"n" is equal to the total numbers u have to choose from

what i've got so far is :
Let x count the number of even sums
(aCm) (bCm)
P(x)= ------------
(a b)Cm

Restrictions to this equation:

a=the amount of even numbers
b=the amount odd numbers
m=the amount of numbers selected
n=the total amount of numbers

when n is odd:
n/2=a
a 1=b

when n is even:
n/2=a
n/2=b

also, this equation only works when m is equial or less then 2. i havent yet figured out how to incorporate how to get the probability of having an even number of odd sums and 1 even numbers.
my equation only covers the probabilities of when the even sum is achieved by either adding all the even number OR all the odd numbers.

the equation works when using these digits:

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

let's say you chose 2=m (so you select any 2 numbers of the 10)

(5C2) (5C2)
P(x)= ----------- =0.444
10C2