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

The number of combinations with 7 unique letters is not 7!, unless you simply are trying to arrange 7 unique tiles that already have been chosen. However, I understood the OP's question to involve 7 unique tiles, chosen from a total of 26 possible letters, which gives a different equation. That equation would be 26 * 25 * 24 * 23 * 22 * 21 * 20. Your first tile could be one of 26 possible letters. Since repeats were excluded, the next tile could only have 25 possibilities - the first chosen letter is "taken". And so on. This equation becomes (A!)/(A-n)!, where the variables are as I described above.

If the blank tile can be *any* letter, including ones that have already been chosen, then you simply multiply by 26, as many times as you've got blank tiles. If the blank tile must also be unique, the it works a though it were just another tile with a letter on it.

When the tiles can all be chosen without regard to whether there is repetition (or when they're all blank), the equation becomes simply A ** n, where A is the size of the alphabet, and n is the number of tiles.

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Re^3: Scrabble word arrangements with blank tiles
by Moriarty (Abbot) on Nov 14, 2005 at 07:00 UTC

    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).