下面的示例比较 Select() 和 SelectMany() 的行为。代码将通过从源集合的每个花卉名称列表中提取前两项来创建一个“花束”。在此示例中，转换函数Select<(Of <(TSource, TResult>)>)(IEnumerable<(Of <(TSource>)>), Func<(Of <(TSource, TResult>)>)) 使用的“单一值”本身就是一个值集合。这需要额外的 foreach（Visual Basic 中为 For Each）循环，以便枚举每个子序列中的每个字符串。

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Text.RegularExpressions;
using System.IO;
using System.Xml;

namespace CSharpTest
{
    class Bouquet
    {
        public List<string> Flowers { get; set; }
    }

class Program
    {
        static void Main(string[] args)
        {
            Console.WriteLine("---------结果------------");
            IEnumerable<int[]> ii = Compute(1, 3, 5, 3);
            foreach (int[] i1 in ii)
            {
                foreach (int i2 in i1)
                    Console.Write(i2);
                Console.WriteLine();
            }

Console.WriteLine("--------string.Concat 连接指定 Object 数组中的元素的 String 表示形式--------");
            Console.WriteLine(string.Concat(new int[] { 1000, 22 }));
            Console.WriteLine("-----------Enumerable.Range 生成指定范围内的整数的序列----------");
            var list = Enumerable.Range(1, 4);
            foreach (var item in list)
            {
                Console.WriteLine(item);
            }
            Console.WriteLine("-------把一个数字转换成 IEnumerable<int[]> 类型--------------");
            ii = from item
                    in list
                 where item == 3
                 select new int[] { item };

foreach (int[] i1 in ii)
            {
                foreach (int i2 in i1)
                    Console.Write(i2);
                Console.WriteLine();
            }

//下面是MSDN的关于select selectmany的例子
            List<Bouquet> bouquets = new List<Bouquet>() 
            {
        new Bouquet { Flowers = new List<string> { "sunflower", "daisy", "daffodil", "larkspur" }},
        new Bouquet{ Flowers = new List<string> { "tulip", "rose", "orchid" }},
        new Bouquet{ Flowers = new List<string> { "gladiolis", "lily", "snapdragon", "aster", "protea" }},
        new Bouquet{ Flowers = new List<string> { "larkspur", "lilac", "iris", "dahlia" }}
    };

// *********** Select ***********            
            IEnumerable<List<string>> query1 = bouquets.Select(bq => bq.Flowers);

// ********* SelectMany *********
            IEnumerable<string> query2 = bouquets.SelectMany(bq => bq.Flowers);

Console.WriteLine("Results by using Select():");
            // Note the extra foreach loop here.
            foreach (IEnumerable<String> collection in query1)
                foreach (string item in collection)
                    Console.WriteLine(item);

Console.WriteLine("/nResults by using SelectMany():");
            foreach (string item in query2)
                Console.WriteLine(item);

}

/// <summary>
        /// 给出sum、min、max和n四个正整数，请输出所有将sum拆分为n个正整数之和，其中每个正整数k都满足：min <= k <= max。这n个正整数之间可以重复，不过由于加法交换率的作用，1 + 2和2 + 1便算是重复的拆分了。
        ///例如，sum = 5，n = 3，min = 1，max = 3，这时候满足条件的拆分方式只有两种：
        /// 1 + 1 + 3 
        /// 1 + 2 + 2
        /// </summary>
        /// <param name="min"></param>
        /// <param name="max"></param>
        /// <param name="sum"></param>
        /// <param name="count">n</param>
        /// <returns></returns>
        public static IEnumerable<int[]> Compute(int min, int max, int sum, int count)
        {
            var list = Enumerable.Range(min, max - min + 1);

if (count == 1)
                return
                    from item
                        in list
                    where item == sum
                    select new int[] { item };

return list.SelectMany
                (
                number => Compute(number, max, sum - number, count - 1).Select
                    (
                    item => item.Concat(new int[] { number }).ToArray()
                    )
                );
        }

}
}