Number of possible moves for the first three steps in checkers/chess.

Question

I'm freshman on StackExchange, so, it's my first question here.

I have a homework in discrete mathematics: I have to count the number of possible moves for the first three steps in checkers and chess.

First, I tried to count it in checkers, because I thought that it is easier. Can you say, did I do that right?

1. For the first move we have $7$ possible ways: $7$,
2. the same for the second: $7\cdot7$ (I multiply it since we have previous moves),
3. and here is the hardest part: I just took a checkerboard and started to count. For the rightmost I have $52$ possible moves and $54 \cdot 2 = 108$ for the others. So, together, it will be $$7 \cdot 7 \cdot (52+54 \cdot 2 \cdot 3)=18324.$$ Am I right?

For chess it was hard too.

1. For the first move we have $20$ possible moves,
2. the same fo the second: $20 \cdot 20$,
3. I find an answer that it is nearly $8000$, but I don't know why. Hope, you will help me.

Thank you!

Answer 1

For checkers/draughts, you are correct that there are $7$ possible first moves, and $7$ possible second moves.

There are two possibilities at this point. The first player had three pieces that could move both left and and right, and one piece that could only move in one direction. If this player moved one of the first kinds of pieces (of which there are $6$ possible moves) in either direction, they have $8$ possible moves for the next turn since they opened up $2$ opportunities below but blocked off $1$ possible move ahead, either by blocking the piece they moved against the edge of the board or by blocking another piece on the row behind. If they moved their one piece that can only move in one direction, though, then they have added $1$ possibility for this piece, $1$ possibility for a piece further back, and blocked one possibility for the piece adjacent, to make a total of $8$ possibilities. So in total we have $6 \cdot 7 \cdot 8 + 1 \cdot 7 \cdot 8$ possibilities for the first three moves. In fact, we can just calculate this as $7 \cdot 7 \cdot 8 = 392$ possibilities, since both of the possibilities for the first move we considered gave an extra possibility in the third move.

The case for chess has been answered here.

Answer 2

My answer that people are linking to counts something very different. For chess, besides the variety of pieces it is complicated because black's move can block white's choice (e4 e5) or offer a new choice (e4 d5). First let us count assuming there are neither of those.
If white moved Na3 or Nh3 there are $14$ pawn moves, $2$ for the other N, and three for the original N for $19$.
If white moves Nc3 or Nf3 there are $14$ pawn moves, $2$ for the other $N$, and $5$ for the moved $N$ for $21$.
If white moves a3 or h3 we lose an N move and a pawn move but gain a R, so there are $19$.
If white moves a4 or h4 there are $21$.
If white moves b3, b4, g3, g4 we add two B moves and lose a pawn move for $21$.
c3 gives three Q moves and loses a P and an N, so $21$.
c4 gives $22$.
d3 adds $5$ for the B and $1$ for the K and Q, less $1$ for the P, $26$
d4 gives $27$.
e3 or e4 gives $4$ for the Q, $5$ for the $B$, $1$ for the K, less $1$ for the $P$, $29$
f3 adds a K loses an N and a P for $19$
f4 gives $20$

If I added right this gives $437$ two move combinations for white assuming no interference from black, so there would be $20 \cdot 437=8740$ three ply starts. We lose $8$ for things like a4 a5 and gain 14 for things like a4 b4 ab, so the total is $8746$

Added: this page cited by Adil Akhmetov gives $8902$ and appears to have done a good computer search. I have not tried to find my error.