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What is the number of ways in which you can reach a point (2,2) from the origin, taking unit steps at a time , in not more than 7 steps?

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You need to have 2 right, and 2 up, and remaining 3 moves have to cancel to 0, so you have 2 right, 2 up, and then either 1 up 1 down, or 1 right 1 left, or nothing at all.

For 6 moves, by symmetry, the correct number is twice the case where you use 1 up and 1 down, and now we're choosing how to order 3 up, 1 down, and 2 right, which should be the multinomial coefficient $$\frac{6!}{3!1!2!}$$So you want twice that, which is 120.

For 4 moves, it's similarly $\frac{4!}{2!2!} = 6$, so you get a total of 126.

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You are assuming (and you are probably right) that these unit steps are to be vertical and horizontal only, though OP never said that. – Gerry Myerson Apr 24 '12 at 13:00
oh yeah good point! I am certainly assuming that. – uncookedfalcon Apr 25 '12 at 0:02

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