How to solve a problem with chessboard I am beginning math student and encountered following task which I do not know how to solve:
There is chessboard, ie 8x8 tiles. In the top left corner there is pawn. He cannot go diagonally and he can only move down and right.
1)How many steps will he need to reach the right bottom corner shortest way?
2)How many of these ways do exist?
Thank you in advance

First you should think - Is there a way which is not the shortest?
Now, He will have to move 7 steps right and 7 steps down. In how many ways can you construct such a path?
EDIT: After having come to the conclusion that the number of such paths is the number of combinations of 14 steps, 7 of which are down and the other 7 of which are right, we reduce the problem to the following: How many sequences of 7 "down"s and 7 "right"s are there?
How do we arrange such a sequence? If we know exactly on which 7 of the 14 steps the pawn went down, then we know that on the rest of the steps (that is, the other 7), he went right. So the problem is equivalent to finding in how many ways we can choose 7 places out of an ordered sequence of 14 places (those chosen ones will be the "down" steps). This number is denoted by $\binom{14}{7}$, or in general if you have $n$ objects out of which you want to choose $k$, then the number of such choices is denoted by $\binom{n}{k}$.
Now let's compute the value of $\binom{14}{7}$. You have 14 options for the first choice (since you can take either one of the 14), 13 options for the second, and so on until you choose the 7th one out of the remaining 8 options. To conclude, this equals to $14\cdot 13\cdot 12\cdot 11\cdot 10\cdot 9\cdot 8$, but then you also have to divide by the number of permutations of the 7th steps, since it is not important whether you, say, take the first step on the first choice and the second on the second choice, or the other way around, so you have $\frac{14!}{7!7!}$ (Why?) - Or in the general case, $\frac{n!}{k!(n-k!)}$.
