Actually, I'm thinking it's a bit more complicated than that. Given seven unique letters, their permutations would be 7!. However, if you're allowed to pick any set of seven (unique) letters out of the alphabet, you would then have a domain of 26 * 25 * 24 * 23 * 22 * 21 * 20 possibilities. That's 3,315,312,000, but who's counting?
With a blank tile, you simply count that tile as one of the letters. The fact that it's blank is something of a red herring. If you're eliminating repeated letters, then the above product would simply have another element - you'd multiply it by 19, because you're just adding another tile, which cannot have any of the previously used letters. Only 19 letters are left. Conversely, if you *replace* one of the letter tiles with a blank one, you simply keep the same product.
On the other hand, if you're *allowing* the blank tile to represent any letter, including repeats, then you multiply your basic rack times 26, as many times as you have blank tiles. So, if your rack is 6 tiles and one blank, your answer is (26!/20!) * 26. Or, with A=size of alphabet, n= number of letter tiles, b = number of blank tiles, you would have:
(A!/(A-n)!) * (b **A) (A ** b).
HTH.
Update: Corrected the order of the operands in the equation.
In reply to Re: Scrabble word arrangements with blank tiles
by spiritway
in thread Scrabble word arrangements with blank tiles
by Anonymous Monk
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