This distribution can be generated using an online calculator and Excel. As an example, suppose we have 19 balls labeled 1-19. Take all combinations of 10 balls and add the numbers on the 10 balls together for each combination. Find the distribution of these sums using the following procedure:
- Use the online combination and permutation calculator
entering 19 for n, 10 for r, no for order, no for repetition, numbers
enter 1 through 19 in list, select cvs
exporting data to excel
There are 92378 combinations for C(19,10).
In excel
calculate the sums for each combination for the data. Use min and max function to determine range
In this example, it ranges from 55 to 145.
Generate the frequencies using the countif function to get the distribution.

To get the mean, use the average function.
To get the standard deviation, use the stdevpa function.
To get the probability mass function, calculate the probability by frequency/total combinations of each unique sum.
In our example, probability of 100 is 2934/92378 = 0.031761 rounded.
The mean can also be calculated using the formula
$\mu = \sum x p(x)$, and standard deviation using $\sigma = \sqrt {\sum (x - \mu)^2 p(x)}$
This example has a mean of 100, and standard deviation of 12.24745 rounded.
The distribution in this example is a symmetric unimodal discrete distribution. The referenced online calculator can generate only a maximum of 1,000,000 combinations.
To get formulas for mean and standard deviation in our problem in terms of a and b,
mean of population $ =
\frac{\sum\limits_{i=1}^n a_i}{a}$
$$= \frac{a(a+1)}{2} * \frac{1}{a}$$
$$= \frac{(a+1)}{2}$$
mean of sample means $\bar x$ (mean of sampling distribution) is same as population mean so
$\frac{(a+1)}{2}$
mean of sampling distribution of sample sums
$\frac{b(a+1)}{2}$
for the standard deviation of the sampling distribution of sample sums, it is
sample size * standard error mean * finite population correction factor
$b * \frac{\sigma}{\sqrt b} * \frac{(a-b)}{(a-1)}$
in our example using the formulas,
mean of the sample sum
$\frac{10(19+1)}{2} = 100$
standard deviation of sample sums =
$10 * \frac{5.47723}{\sqrt{10}} * \frac{19-10}{19-1}$
= 12.24745 rounded
which are the same values we got above with generating the distribution
The distribution is the probability mass function which has no single formula, but can be calculated for each discrete random variable by using variations of dice sum distributions (with replacement) b number of a-sided die, or subset sum (without replacement) specific sized subset calculations for the numerator, and ${{a}\choose{b}}$ for the denominator. The range of the random variables is the integer sequence from $\sum\limits_{i=1}^b x_i$ to $\sum\limits_{i=a-b}^a x_i$.
with replacement, multiple dice sum distributions
http://mathworld.wolfram.com/Dice.html
without replacement, special case of the subset-sum problem in which the number of items in the subset is b
https://mathoverflow.net/questions/101971/subset-sum-problem-when-the-number-of-integers-in-the-sum-is-known
https://math.stackexchange.com/questions/953908/probability-that-the-sum-of-three-integer-numbers-from-1-to-100-is-more-than-1/953950#953950