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Andrew Jull writes: >Assuming a block size of 6 to randomise equivalent numbers >to three different treatment groups (A,B,C), how do I find out >what the total different number of combinations of AABBCC >etc are, short of working it out by hand? I like difficult problems like these. It's job security for statisticians. Some people consider this stuff boring. I can't imagine why. If you're one of those people, please delete this email. The answer is 90. There are a couple of approaches that work. Let C(n,k) be the combination formula, that is, C(n,k)=n!/(k!(n-k)!) There are 2 A's and six positions to place them in. So the number of ways to place the As is C(6,4). There are 2B's and four remaining positions to place them in. The number of ways to do this is C(4,2). Once the A's and B's are in place, there is only one way to position the C's. Using the multiplication rule, there are C(6,4)*C(4,2)*1 ways to place the A's, B's and C's. This works out to equal 15*6*1=90. Another approach is to note that there are 6!=720 ways to arrange 6 items. But half (720/2=360) of these arrangements are redundant, because the A's are indistinguishable. Another half (360/2=180) are redundant because the B's are indistinguishable. Still another half (180/2=90) are redundant because the C's are indistinguishable. Whew! There's a way to do block randomization with having to compute and tabulate these combinations. See http://www.childrens-mercy.org/stats/plan/random.asp Steve Simon, [log in to unmask], Standard Disclaimer. STATS: STeve's Attempt to Teach Statistics. http://www.cmh.edu/stats Watch for a change in servers. On or around June 2001, this page will move to http://www.childrens-mercy.org/stats