Some creative nth-child()-ing. In a table-esque inline-block list, it will remove the extra line that results from needing both container and list item borders.
Why might this be useful?
For this example, I’ll be using a 3-column layout using only inline-block list items. The three selectors used in combination to select the bottom row in all permutations are:

For this example, I’ll be using a 3-column layout using only inline-block list items. The three selectors used in combination to select the bottom row in all permutations are:

There’s a pattern/strategy here. Let x be the total number of columns. All we have to do is figure out how to select each list item when it’s on the bottom row, and deselect it when it’s not. Let’s try this with four columns:
li:last-child
li:nth-last-child(2):not(:nth-child(xn))
li:nth-last-child(3):not(:nth-child(xn)):not(:nth-child(xn-1))
li:nth-last-child(4):not(:nth-child(xn)):not(:nth-child(xn-1)):not(:nth-child(xn-2))
For additional columns, add another selector incrementing the nth-last-child() parameter, and add another :not(:nth-child(xn-?)), incrementing the ?.
Questions? Find me on Twitter and ask away! Or just view some working code on Dabblet.
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If I had only one hour to save the world, I would spend fifty-five minutes defining the problem, and only five minutes finding the solution. — Albert Einstein