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

In all cases, the total number of sums is the number of combinations of m elements out of n elements which is mCn.

You may choose m even numbers and 0 odd numbers to the total of mCa·0Cb, or you may choose (m-2) even numbers and 2 odd numbers to the total of (m-2)Ca·2Cb, or (m-4) even numbers and 4 odd numbers to the total of (m-4)Ca·4Cb, and so on. You'll have to add up all the products and divide the result by mCn. Can't tell you off the top of my hat whether the result may be given by a simple formula. (I assumed rCs = 0 wherever r > s.)

I would not concentrate on numbers of combinations, when talking about even and odd numbers we know that

odd + odd = even
odd + even = odd
even + odd = odd
even + even = even

This can be generalised to

sum( any number of even numbers ) = even
sum( any number of even numbers plus an even number of odd numbers) = even
sum( any number of even numbers plus an odd number of odd numbers) = odd

So the question can be restated as I pick m integers from (1, 2, 3, ..., n ) what is the probability that I have an odd number of odd numbers.

Without telling you the answer consider the cases m=1, m=2 (shown above), m=3, this will give you a clue to the answer without proving it however from these results generalise a solution of m+1 to prove the answer.

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???