I even don’t know where I heard of this question but I’ve been thinking about it and I’m not sure how I should approach this.

Consider a chess board (8*8) and the king in leftest and downest squire. This king wants to go to the rightest and upest square . this king is only interested in moving toward his goal , so if he wants to move from his place he always choose between going “RIGHT” “UP” "NORTH EAST (on diagonal)” .there is no coming back. He wants to go to rightest and upest square as we mentioned. How can we calculate possible ways of him doing it ?

At first I tried to estimate the longest way possible (which is 14) if we just use our right and left option . Moving on diagonal is a faster approach.every move on diagonal take two moves if we just use left and right . Everytime we use an extra diagonal move we reduce two right or left moves So we if we choose

0 diagonal move we have 14 right or left

1 diagonal moves we have 12 right or left

2 diagonal moves we have 10 right or left



7diagonal moves we have 0 right or left

If I want to calculate the first one with 0 diagonal move and 14 right and left logically I have 7 rights and 7 lefts to use but if my first move is right i have 7 possible ways to go up and if i take two rights at first I have 7 absolutely different ups to go. How should I even write this in combination language? If we use 1 diagonal move it’s even worse , how can we calculate where in the road should we do that ?

I’m really lost Please answer or help me answer it.

Thank you guys

  • $\begingroup$ try it with 3 by 3 and 4 by squares, the answer should be obvious for n by n. $\endgroup$
    – jimjim
    Jun 9 '20 at 5:18
  • $\begingroup$ Try it with a smaller square. Not to stumble on a number and make a guess, but to recognize clearly how so efficient systemize it. $\endgroup$
    – fleablood
    Jun 9 '20 at 5:27

If you are not allowed to use any diagonal moves, then the number of ways is just $\frac{14!}{7! 7!}$, since any combination of $RRRRRRR UUUUUUU$ ($7$ right, $7$ up) works.

If you can use exactly one diagonal move, note that this move replaces one pair of $RU$. There are now $13$ possible moves, which include $6$ R's, $6$ U's, and one diagonal move. This gives $\frac{13!}{6! 6! 1!}$ combinations.

If you can use exactly two diagonal moves, then similarly, the number of combinations is $\frac{12!}{5! 5! 2!}$.

Can you continue from here?

  • $\begingroup$ I don’t get why $\frac{14!}{7!7!}$ is the number of ways without any diagonal move. Can you explain more ? And explaining just one would be enough.thank you $\endgroup$
    – Negar
    Jun 9 '20 at 5:06
  • $\begingroup$ You can choose the $R$'s first, of which there are $7$ of them. However, since you cannot tell if you move one $R$ in the sequence to a different place (the $R$'s are indistinguishable), you have to divide by $7!$. Similarly, for the $U$'s, you have to divide by $7!$ as well. This number is also $14$ choose $7$. $\endgroup$
    – Toby Mak
    Jun 9 '20 at 5:11
  • $\begingroup$ Thank you alot <3 $\endgroup$
    – Negar
    Jun 9 '20 at 6:08

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