Anna and Ben live at opposite corners of a \(m \times n \) block.

If Anna only moves right or up or down, and isn't allowed to retrace her steps, how many ways does she have to get to Ben?

If Anna only moves right or up or down, and isn't allowed to retrace her steps, how many ways does she have to get to Ben?

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TopNewestm^(n-1)

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Why?

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(m+1)^n it is?

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Comment deleted Aug 19, 2015

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How would you generalize it to an \(m \times n \) board? And why is that the answer?

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Start at the beginning. Anna can move right or down. If she moves down, she can only move down until she moves right. She can move to the second column in n+1 ways. The same case occurs for every column. Anna will have n+1 ways of moving to the next column each time. This keeps going until she is in the last column, in which case she has to move toward Ben. There are m+1 columns, each of which give n+1 ways of moving to the right, except for the last column. Therefore there m^(n+1) ways for Anna to reach Ben.

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Yes, that's the quick way to recognize it. Once you choose the horizontal paths (one in each column), then the vertical paths are fixed. This is why we can set up the bijection.

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Comment deleted Aug 03, 2015

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Close, but not quite. Look at the other parts, and see if the answer matches your guess of \( m+n \choose n \).

You are forgetting that Anna is allowed to move up

anddown (but not retrace her steps).Log in to reply

There are (m+1) ways to enter each column. Anna has to enter n such columns consecutively. So the number of ways is (m+1)(m+1).....upto n times, i. e. (m+1)^n. This comes up because she is not allowed to go left. If she could go right, left and down, answer would be (n+1)^m with (n+1) ways to enter each of the m rows (WLOG).

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