Get a record position in a group + record count of the group on each record

Question

I have a resultset in a temp table on SQL Server 2008. One of the fields is a group Id, and for each group I can have any number of rows. I'd like to get in a new column the position of the row in the group, such as "row 1 of 3" So, if I have temp table #Data(GroupId int, Item char) with

GroupId  Item
-------------
   5      C
   5      A
   5      T
   2      U
   3      Y
   3      L

I'd like to get

GroupId  Item Pos GroupCount
----------------------------
   5      C    1     3
   5      A    2     3
   5      T    3     3
   2      U    1     1
   3      Y    1     2
   3      L    1     2

I managed to do this doing

SELECT D.GroupId, D.Item, ROW_NUMBER() OVER (PARTITION BY D.GroupId ORDER BY D.Item) AS Pos, COUNT(1) AS GroupCount
  FROM #Data D
       INNER JOIN #Data DC
 GROUP BY D.GroupId, D.Item

But I'm not too happy with this, because the query is actually a little more complex because of all the columns I actually have to retrieve, and because doing both an inner join with itself to get the group row count and a partition to get the row position seems like too much. Is there a simpler way to do this?

---

I didn't exactly use Tim's answer, but it showed me the way:

SELECT D.GroupId, D.Item,
       ROW_NUMBER() OVER (PARTITION BY D.GroupId ORDER BY D.Item) AS Pos,
       COUNT(D.GroupId) OVER (PARTITION BY D.GroupId) AS GroupCount
  FROM #Data D

Answer 1

You can use COUNT(GroupID)OVER(PARTITION BY GroupID):

WITH CTE AS
(
  SELECT GroupId, Item
  , GroupCount = COUNT(GroupID) OVER (PARTITION BY GroupID)
  , Pos = ROW_NUMBER() OVER (Partition By GroupID Order By GroupId)
  FROM #Data
)
SELECT GroupID, Item, Pos, GroupCount 
FROM CTE

OVER Clause (Transact-SQL)

Here's a fiddle: http://sqlfiddle.com/#!6/b6505/2/0

Note that it's not ordered since you haven't told how to order.