excel - Differentiating between cells that have the same information for match function -

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lets have these 3 tables show interest rates various different things (e.g. auto loan, mortage, credit cards). "######"s showing there values in cells used calculate numbers @ bottom (0.01, 0.03, etc.). lets range in excel these 3 data tables a1:i6.

|              |   datatable 1   |               |              |   datatable 2   |               |              |   datatable 3   |               | |:------------:|:---------------:|:-------------:|:------------:|:---------------:|:-------------:|:------------:|:---------------:|:-------------:| | low,interest | medium,interest | high,interest | low,interest | medium,interest | high,interest | low,interest | medium,interest | high,interest | |--------------|-----------------|---------------|--------------|-----------------|---------------|:-------------|-----------------|---------------| |    #######   |     #######     |    #######    |    #######   |     #######     |    #######    |    #######   |     #######     |    #######    | |    #######   |     #######     |    #######    |    #######   |     #######     |    #######    |    #######   |     #######     |    #######    | |     0.01     |       0.03      |      0.05     |     0.02     |       0.04      |      0.06     |     0.10     |       0.20      |      0.30     |

i have drop down list in a8 contains values data table 1, data table 2, , data table 3.

lets have table (range k1:m14).

|   month   | balance | medium interest | |:---------:|:-------:|:---------------:| |  january  | $100.00 |        3%       | |  february | $103.00 |        3%       | |   march   | $106.09 |        3%       | |   april   | $109.27 |        3%       | |    may    | $112.55 |        3%       | |    june   | $115.93 |        3%       | |    july   | $119.41 |        3%       | |   august  | $122.99 |        3%       | | september | $126.68 |        3%       | |  october  | $130.48 |        3%       | |  november | $134.39 |        3%       | |  december | $138.42 |        3%       |

i wrote formula determine how 3% gets medium interest column.

=index($a$6:$i$6,match($a$8,$a$1:$i$1,0),match($m$2,$a$2:$i$2,0))

it works when choose data table 1 in drop down list. correctly places 3% medium interest rate data table 1 when choose either of other 2 data tables, invalid cell reference error. appear me problem formula cannot differentiate between low, medium, , high columns different tables.

this need in real spreadsheet.

please try: 

=index($a$6:$i$6,match($a$8,$a$1:$i$1,0)+match($m$2,$a$2:$c$2,0)-1)

but note matches must exact, in particular label columnm , corresponding indicators in columnsa:i.

the first match provides index number position of value selected drop-down because data in blocks identifies of 3 blocks. in addition necessary identify column block, achieved matching columnm label within first block - return 1 (low), 2 (medium) or 3 (high). since each block starts low, adding 1 (for low) , subtracting 1 provides offset required other match find anyway (if first column of whichever block) + 0 still first column (low) of whichever block. 2 (medium) -1 1 index function takes 1 more index number returned first match.


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