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But, if you have 1 parity set of 24 drives, you can only lose
1 and keep running, correct? With 3 parity sets of 8 drives,
you could theoretically lose 3 drives and still be running. I
say "theoretically" because the 3 failures would have to be 1
drive in each of the 3 parity sets as opposed to any 3 random
drives.
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