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In permuted block randomization, a block size of $4$ has $6$ possible permutations:

$AABB, ABAB, ABBA, BBAA, BABA, and BAAB$

And a block size of 6 has 8 possible permutations:

$AAABBB, BBBAAA, AABBAB, BBAABA, ABABAB, BABABA, ABAABB,$ and $BABBAA$

What formula is used to generate these permutations? Each block must contain an equal number of $As$ and $Bs$. Why isn't $ABBAAB$ a possible permutation when using a block size of $6$?

Amzoti
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SEL
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    These slides (see p.16) and this paper (p.347, the end of the first paragraph of the "Methods" section) say that all ${6\choose3}=20$ permutations can be used, but a number of documents on the internet do say that there are only eight possible permuations for a block size of $6$. So I guess the reason is not purely mathematical but something related to the context of the problem. – user1551 Feb 12 '13 at 19:15

1 Answers1

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They don't explain it. For a block size of $6$ with $3\ A$'s, there are ${6 \choose 3}=20$ possibilities. It sounds like they thought this was "random enough".

Ross Millikan
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