Counting pairs of elements

//The pattern conveys - the first digit followed by d other digits then followed by the second digit
//firstNo="1"; secondNo="2"; distance="2"; All firstNo,secondNo,distance are String.
Regex regexObj=new Regex(firstNo + @"\d{" + distance + "}" + secondNo);
//Match collection gets all occurences of this regular expression
MatchCollection MatchObj = regexObj.Matches("1122112213333344");
//Just printing the occurence count!
Console.WriteLine(MatchObj.Count);

int[] arr = {2, 1, 1, 2, 2, 1, 3, 3, 3, 3, 3, 4, 4};
int d = 2;
int a = 1
int b = 2;
var e = arr.Skip(d).GetEnumerator();
int m = arr.Take(arr.Length-d).Count(v=>e.MoveNext() && v==a && e.Current==b);

static void Main(string[] args)
{
    int[] arr = {2, 1, 1, 2, 2, 1, 3, 3, 3, 3, 3, 4, 4};
    int step = 2;

    var r = GetPermutations(arr, step);
    Console.WriteLine("arr  = {0}", string.Join(",", arr));
    Console.WriteLine("step = {0}", step);
    foreach (var e in r)
    {
        Console.WriteLine("{0}-{1} = {2}", e.Key.Item1, e.Key.Item2, e.Value);
    }

}
// returns unsorted dictionary of permutation counts of pairs step index positions appart
public static Dictionary<Tuple<int, int>, int> GetPermutations(int[] arr, int step)
{
    Dictionary<Tuple<int, int>, int> r = new Dictionary<Tuple<int, int>, int>();
    var e1 = arr.Take(arr.Length-step).GetEnumerator();
    var e2 = arr.Skip(step).GetEnumerator();
    while(e1.MoveNext() & e2.MoveNext())
    {
        Tuple<int, int> k = new Tuple<int,int>(e1.Current, e2.Current);
        if (r.ContainsKey(k)) r[k]++;
        else r.Add(k, 1);
    }
    return r;
}

arr  = 2,1,1,2,2,1,3,3,3,3,3,4,4
step = 2
2-1 = 2
1-2 = 2
2-3 = 1
1-3 = 1
3-3 = 3
3-4 = 2

static void Main(string[] args)
{
    int[] arr = {2, 1, 1, 2, 2, 1, 3, 3, 3, 3, 3, 4, 4};
    int step = 2;

    var r = GetPermutations(arr, step);
    Console.WriteLine("arr  = {0}", 
               string.Join(",",
                      Array.ConvertAll(arr, 
                           delegate(int i) { return i.ToString(); })));
    Console.WriteLine("step = {0}", step);
    foreach (var e in r)
    {
        Console.WriteLine("{0}-{1} = {2}", e.Key.Key, e.Key.Value, e.Value);
    }

}
// returns unsorted dictionary of permutation counts of pairs step appart
public static Dictionary<KeyValuePair<int,int>,int> GetPermutations(int[] arr, int step)
{
    Dictionary<KeyValuePair<int,int>,int> r = 
                  new Dictionary<KeyValuePair<int,int>,int>();
    var e1 = arr.GetEnumerator();
    var e2 = arr.GetEnumerator();
    while ((step > 0) && e2.MoveNext()) step--;
    while(e1.MoveNext() & e2.MoveNext())
    {
        KeyValuePair<int, int> k =
               new KeyValuePair<int, int>((int)e1.Current, (int)e2.Current);
        if (r.ContainsKey(k)) r[k]++;
        else r.Add(k, 1);
    }
    return r;
}