# Max pieces to cover a chess board [closed]

Number of individual chess pieces to cover all chess board

## closed as unclear what you're asking by Rubio♦Apr 18 at 23:15

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

• Are they three identical chess pieces? – Weather Vane Apr 18 at 20:10
• What do you mean by "possible positions"? Legal positions that could happen in a real game? If so, 2 of the 3 pieces have to be kings. If not, then why does it matter at all what the 3 pieces are? – GendoIkari Apr 18 at 20:11
• Or, do you simply mean to find the number of ways you can place 3 objects on an 8x8 grid so that none are in the same rank or file as another object; nothing to do with actual chess piece rules? – GendoIkari Apr 18 at 20:14
• Or it could essentially how many ways can 3 objects be arranged on an 8x8 grid without being on the same rank, file, or diagnol, and chess is simply being used as analogy. – Rewan Demontay Apr 18 at 20:18
• Closing this until the asker's intent has been made clear. We've already had one completely different version of this question answered, an answer which now looks comically misplaced because the current question is completely different (and is now almost meaningless). Certifill, please figure out what ONE question you're trying to ask, and ask it clearly and completely. – Rubio Apr 18 at 23:17

Making an assumption that the intent is to find the number of different ways that 3 rooks could be placed on a chess board without attacking one another. In other words, not the "maximum number", which implies trying to find the right 3 pieces, but simply the number of different ways.

Brute forcing it:

The first piece has

64 different places where it can be placed.

For each of those options, it

Eliminates 15 total places where the second piece can be (its entire row, and its entire column, subtracting 1 for the square that would be counted twice. Leaving the second piece with 49 spaces to be.

For each of those options, it

Eliminates another 13 total spaces... 2 fewer because of double-counting with the spaces eliminated by the first piece. Leaving (64-15-13) = 36 spaces for the final piece.

This gives us a total of

64*49*36 = 112896 different arrangements. But because the 3 pieces are assumed to be identical, we need to divide by 3!=6