In how many different ways can you place the black rook and the black king if they must be next to each other (in any direction including diagonally). The two first diagrams show acceptable positions whereas the third one is not acceptable because there is at least one empty square between the rook and the king.

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Here is what I have so far:

Case 1

Black king in the corners-> 4 possible choices

Case 2

Black king on the sides not a corner->24 possible choices

Case 3

Black king on the interior->30 possible choices

I want to know if I am on the right track?

  • $\begingroup$ Place the King (corner, edge or interior) and then count the possibilities for the rook $\endgroup$
    – Henry
    Apr 30 '17 at 19:02

You are headed in the right direction.

Case 1: King in a corner. There are 4 corner squaress for the king, and for each square, the rook has 3 positions, giving $4 \cdot 3 = 12$.

Case 2: King on an edge. There are 24 edge squares for the king, and for each square, the rook has 5 positions, giving $24 \cdot 5 = 120$.

Case 3: King in the center. There are 36 center squares for the king, and for each square, the rook has 8 positions, giving $36 \cdot 8 = 288$.

These cases are exclusive (they do not overlap) and they are exhaustive (all possibilities are covered). Hence, adding them all up we get $12 + 120 + 288 = 420$.

Note that this method can be easily generalized to placing a rook and king on an $n \times n$ chessboard, or even on a $k$-dimensional chessboard. With some effort, you can place more pieces, etc.

  • $\begingroup$ I am not following case 2 $\endgroup$
    – Phillip Mc
    Apr 30 '17 at 19:30
  • $\begingroup$ @PhillipMc If you place the king along an edge, but not in a corner, then there should be 5 spaces for the rook. For example, if the king is along the left edge, then the rook can be 1) above, 2) below, 3) right, 4) diagonally above-right, and 5) diagonally below-right from the king. By symmetry, you always get 5 possible positions for the rook when the king is along en edge not in the corner. $\endgroup$
    – Bram28
    Apr 30 '17 at 19:53

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