How to test whether a subset is representative of a population? Given that I have an entire population's data, I'm looking to see if a subset that I take is representative of the population, on average. I'm not choosing randomly, as I plan to pick each item exactly one time and add it to a subset. In other words, I plan to break the population into chunks of size N, where each is representative of the population, using each item from the original population exactly once. Is there a sound way to test that each subset is representative?
Example
Given the population [0, 0, 0, 0, 0, 0, 0, 0, 1, 1], I want to get subsets of size N=5 that are representative of the full population. I then generate the subsets [0, 0, 0, 0, 1] and [0, 0, 0, 0, 1]. In this trivial case, I can easily test that the percentage in the population matches the percentage in the subset. Doing so we'd see that they're perfect representations of the population.
However, say the population is instead [0, 0, 0, 0, 1, 1, 1, 2, 2, 3]. Assuming that I again want to get subsets of size N=5, I'd generate [0, 0, 1, 1, 2] and [0, 0, 1, 2, 3]. In this case, each subset is not a perfect representation of the population — the first subset doesn't even represent 3. These examples are small for the sake of explanation, but I'm wondering if there's a reasonable way to test that, on average, the stratified subsets are representative of the population.
(For the curious: I have written some code to generate the subsets, and I'm looking to verify that my implementation is correct. If there's a standard way to generate them, by all means please let me know.)
 A: If I correctly understand what you're doing, on average each chunk will 'represent' the population in that each observation has a fair chance of being included, and $E(\bar X_{subset})=E(\bar X_{whole})= \mu_{population}.$ (And similarly for variances.) But you can't expect all chunks to $exactly$ match the composition of the sample. 
Random subsets can be obtained by making a random permutation of the data, and then partitioning it into subsets. However, your reason for wanting such subsets is unclear.
Below I have taken a chunk and compared its mean and variance with the mean and variance of the whole sample. P-values > 5% indicate 
no differences detected.
 n = 500;  s = rnorm(500, 100, 15)  # sample of 500 from normal
 perm = sample(s, n)  # random permutation of sample
 chunk = perm[1:250]  # first 250 observations in permuted sample

 var.test(s, chunk)   # tests whether s and chunk have = variances

 ##        F test to compare two variances

 ## data:  s and chunk 
 ## F = 0.9659, num df = 499, denom df = 249, p-value = 0.7425
 ## alternative hypothesis: true ratio of variances is not equal to 1 
 ## ...

 t.test(s, chunk)

 ##   Welch Two Sample t-test  # tests whether s and chunk have = means

 ## data:  s and chunk 
 ## t = 0.6466, df = 490.505, p-value = 0.5182
 ## alternative hypothesis: true difference in means is not equal to 0 
 ## ...

A Kolmogorov-Smirnov test shows good fit of the chunk with
the normal distribution used to simulate the the data.
A K-s test can also compare two samples, but not a sample and
a subsample of it (on account of the ties produced).
 ks.test(chunk, pnorm, 100, 15)  # tests whether chunk agrees w/ pop

 ##   One-sample Kolmogorov-Smirnov test

 ## data:  chunk 
 ## D = 0.0739, p-value = 0.1304
 ## alternative hypothesis: two.sided 

However, for continuous data (no ties) a K-S test can
check if two non-overlapping chunks come from the same distribution.
Again here, no discrepancy is detected.
 chunk2 = perm[251:500]  # second chunk
 ks.test(chunk, chunk2)  # tests whether two chunks from same population

 ##   Two-sample Kolmogorov-Smirnov test

 ## data:  chunk and chunk2 
 ## D = 0.072, p-value = 0.5361
 ## alternative hypothesis: two.sided 

