I have a table set up in excel as follows:

Town 1 2 3 4
Building 1 $12MM $12MM $8.5MM $8.5
Type 2 $10MM $10MM $6.5MM $6.5MM
3 $7.5MM $7.5MM $5.0MM $5.0MM
4 $3.0MM $3.0MM $1.0MM $1.0MM
5 $1.5MM $1.5MM $500M $500M

I've created the following formula where F25 is "town" and F24 is "building
type.

"=INDEX($Z$24:$AJ$35,MATCH(F25,$Z$24:$Z$35,),MATCH F24,$Z$24:$AJ$24,))".

I need to add a third criteria - F23 - which is "quality" - denoted by 2, 3
or 4. If it is 2, the formula noted above would work. However, if it is 3
or 4, the values in the table would change; for example, if the "quality" was
3, the "town" is 1 and the "building" is 1, the value returned would be $10.5
MM. If it was 4, the value returned would be $5MM.

I have created two additional tables that mirror the table noted above with
the exception that the values shown in each column are different; I've copied
the formula noted above into the new tables, amending the cell references to
that table (ie. @index(X45:AJ57, MATCH(F25,X45:Z57,) etc., etc. I've then
set up an @if statement which reads
"=IF(F23=2,Z37),=IF(f23=3,Z59),=if(f23=4,Z81)) )". If I type in 2 in the
cell, the value returns. If I type in 3, I get "FALSE". The same with 4.
I'm stuck. I know it has something to do with the fact that there is nothing
pointing to the other two tables in the IF statement. Do I name the tables -
eg. RG2, RG3, RG4? If that's the route, can I use an IF statement? If so,
how should it read?
IF(F23=2,RG2,Z37),=IF(F23=3,RG2,Z62),IF(f23=4,RG4, Z83)))? That doesn't seem
to work - I get "your formula contains an error - check Excel Help". Help
doesn't "help" in this case. I tried to jerry rig the @index formula to
include the third criteria, but I got the FALSE return as well. I tried to
create one table with all values, but it's too big and confusing.

I would really appreciate any help you can provide. I'm sorry this is so
long winded, but I've tried to include as much detail as I could to outline
the problem.

Thanks much.

Marg