Get next N elements from enumerable

Question

Context: C# 3.0, .Net 3.5
Suppose I have a method that generates random numbers (forever):

private static IEnumerable<int> RandomNumberGenerator() {
    while (true) yield return GenerateRandomNumber(0, 100);
}

I need to group those numbers in groups of 10, so I would like something like:

foreach (IEnumerable<int> group in RandomNumberGenerator().Slice(10)) {
    Assert.That(group.Count() == 10);
}

I have defined Slice method, but I feel there should be one already defined. Here is my Slice method, just for reference:

    private static IEnumerable<T[]> Slice<T>(IEnumerable<T> enumerable, int size) {
        var result = new List<T>(size);
        foreach (var item in enumerable) {
            result.Add(item);
            if (result.Count == size) {
                yield return result.ToArray();
                result.Clear();
            }
        }
    }

Question: is there an easier way to accomplish what I'm trying to do? Perhaps Linq?

Note: above example is a simplification, in my program I have an Iterator that scans given matrix in a non-linear fashion.

EDIT: Why Skip+Take is no good.

Effectively what I want is:

var group1 = RandomNumberGenerator().Skip(0).Take(10);
var group2 = RandomNumberGenerator().Skip(10).Take(10);
var group3 = RandomNumberGenerator().Skip(20).Take(10);
var group4 = RandomNumberGenerator().Skip(30).Take(10);

without the overhead of regenerating number (10+20+30+40) times. I need a solution that will generate exactly 40 numbers and break those in 4 groups by 10.

Answer 1

Are Skip and Take of any use to you?

Use a combination of the two in a loop to get what you want.

So,

list.Skip(10).Take(10);

Skips the first 10 records and then takes the next 10.