Jim: Thanks for the help: I changed the variable type to interger and the
new code with cells worked! Again Thanks!

"Jim Thomlinson" wrote:

Give this a try...

Cells(IngRowsxy, columnxy).Select

Range takes up to 2 arguments. Each argument is a cell or cell address. Your
code supplies the references to a single cell. You want to use Cells for
that...
--
HTH...

Jim Thomlinson


"Mathew" wrote:

Gary's Student: Tried this and it did not work. I also changed the variable
names. Here is what I have: I have a variable named IngRowsxy which equals
an integer between 13 and 31 depending on several other options selected.
The macro is assigning the correct value to this variable. The other
variable is columnxy which is either 4 or 5. This is also working correctly.
I used the step-by-step execution of the macro to verify this. However,
when I try to use the Range(columnxy, IngRowsxy).Select I get the error. I
also tried this: Range(columnxy IngRowsxy).Select So with your code,
below I tried:


addrs = IngRowsxy & "," & columnxy
Range(addrs).Select

and when that failed I tried:

addrs = IngRowsxy & columnxy
Range(addrs).Select

Neither of these worked! Any help would be appreciated.