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Re^3: Scrabble word arrangements with blank tiles

by Moriarty (Abbot)
on Nov 14, 2005 at 07:00 UTC ( #508223=note: print w/replies, xml ) Need Help??


in reply to Re^2: Scrabble word arrangements with blank tiles
in thread Scrabble word arrangements with blank tiles

Actually, if I'm reading the OP correctly, they already have the 7 unique tiles and want to know how many possible combinations they can make from them, which would be 7!

If you replace one with a blank, the number of combinations depends on the rules for the blank. If the blank can only be one of the remaining letters, then the number of combinations becomes 7! * 20. If, however, the blank can be any letter, including any of the other 6 chosen, the formula becomes more complex, as you would have 7! * 26 minus any duplicates (I haven't worked out how to calculate the number of duplicates that would be involved).

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