How many ways are there for a piece in the bottom-left corner of a 5x5 chessboard to move to the square marked $B$ in the figure below if the piece may only move up, right, and diagonally to the upper-right one square at a time? enter image description here

Attempt: I imagined the board as closed rooms and in each room I placed a door connecting each two rooms. I counted the doors, eliminated the repetitions and gave 75. As I could repeat the path in a perpendicular direction from the first room I added another 75. I Got $76+76=152$

I would like to know the flaw in my reasoning... the answer to the question is 321


1 Answer 1


I don't get really well what you have calculated but here is how I would do it : use the fact that anytime you use a diagonal move you consume one up and one right.

You can achieve the top right with $0$ diagonal move and $8$ moves, $4$ right, and $4$ up moves in any configuration, or $1$ diagonal moves and $6$ moves, $3$ right, and $3$ up moves in any configuration, etc.

So you have that the total combinations are

$${8 \choose 4} + {7 \choose 1}{6 \choose 3 }+ {6 \choose 2} {4 \choose 2}+ {5\choose 3}{2 \choose 1 }+ { 4 \choose 4} = 321$$

  • 2
    $\begingroup$ If you make three diagonal moves, there are a total of five moves, of which three are diagonal, one is up, and one is to the right. Therefore, the fourth term in your sum should be $\binom{5}{3}\binom{2}{1}$ since you must select which three of the five moves are diagonal moves and which of the other two moves is to the right. $\endgroup$ Oct 15, 2021 at 22:48
  • $\begingroup$ yes totally right I was trying to figure out why It was broken thank you, somehow I lost the logic in the middle :-) $\endgroup$
    – Tortar
    Oct 15, 2021 at 22:49

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