Combination Formula with a lockout twist

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To calculate the number of combinations for selecting one item from each of six groups of three, where selecting one item locks out the others in that group, the formula involves understanding the concept of combinations with restrictions. Each group allows for three choices, but only one can be selected, leading to a total of 3^6 combinations, equating to 729 possible selections. The "lockout" condition ensures that only one item from each group is chosen, simplifying the calculation. To clarify the setup, starting with smaller examples can help visualize the process before applying it to the larger problem. The final goal is to determine the total combinations while adhering to the lockout rule.
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Combination Formula with a "lockout" twist

Hi! I am trying to figure out all possible combonations for 6 items among a group of 18 choices. So I turn to my old friend C(n,r) to calculate where n=18 and r=6. "But WAIT!" I tell you before you hastily begin scribbling, "There is a twist..." You see my problem is that the items are divided up into 6 groups, with 3 choices in each group. Once a choice has been made in a group for the combination the other 2 in the group are unavailable, or "locked out" of the rest of the combination. The order doesn't necessarily matter but a choice must be selected from each of the six groups. Here's a visual representation:

A B C
1 A1 B1 C1
2 A2 B2 C2
3 A3 B3 C3
4 A4 B4 C4
5 A5 B5 C5
6 A6 B6 C6

If "B1" is selected in a single combination then "A1" and "C1" cannot be apart of the same combination. What is the formula for this and how many possible combinations are there?
 
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Try to start with a smaller problem (1,3) and (2,6) which you can count and then calculate this one. Honestly, I didn't quite understand the setup.
 

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