Please check the attachment for the figure. enter image description here In how many ways, can a person go from point A to point B if he or she could only move down or right?

My answer is 9!/(5!4!).

Please let me know if I am right or wrong? And correct method too?

  • $\begingroup$ My answer is 126 ways. Definitely more than 1 ways. $\endgroup$ Jan 2, 2019 at 13:42
  • $\begingroup$ No. In a straight line, he could move either right or down. Don’t think of it as a whole block. $\endgroup$ Jan 2, 2019 at 13:48

2 Answers 2


This diagram should make it clear that there are $29$ paths. \begin{array}{rrrrrrrrr} \text{A}&\rightarrow&1&\rightarrow&1\\ \ \downarrow&&\ \downarrow&&\ \downarrow\\ 1&\rightarrow&2&\rightarrow&3\\ &&\ \downarrow&&\ \downarrow\\ &&2&\rightarrow&5\\ &&\ \downarrow&&\ \downarrow\\ &&2&\rightarrow&7\\ &&\ \downarrow&&\ \downarrow\\ &&2&\rightarrow&9&\rightarrow&9&\rightarrow&9\\ &&\ \downarrow&&\ \downarrow&&\ \downarrow&&\ \downarrow\\ &&2&\rightarrow&11&\rightarrow&20&\rightarrow&\text{B} \end{array}


How many ways from A to B in the figure below?

enter image description here

Now in your picture, are there more ways, fewer, or the same number of ways? Does this picture have paths that yours does not? Does your picture have paths that this one does not?

The general formula when you go from one corner of a rectangular grid to the diagonally opposite corner is $\binom{W+H}{W} = \frac{(W+H)!}{W!H!}.$ If you delete some of the lines in the grid, you lose the paths that go on those lines, and you need to do something more complicated to count the paths. Usually this will involve multiple cases, perhaps even inclusion-exclusion.

  • $\begingroup$ m.youtube.com/watch?v=f5AORqe1ADk $\endgroup$ Jan 2, 2019 at 14:23
  • $\begingroup$ I know the answers to all my questions. Have you thought about what I said? $\endgroup$
    – David K
    Jan 2, 2019 at 14:25
  • $\begingroup$ Based on this, 126 ways. Same number of ways. But my approch might be wrong to get 126 ways in my original question. $\endgroup$ Jan 2, 2019 at 14:26
  • $\begingroup$ Your picture is different, but you say the number of paths is the same. How do you figure that? $\endgroup$
    – David K
    Jan 2, 2019 at 14:26
  • $\begingroup$ I don’t know the number of ways for my original question. I was just curious to modify the question given in the video. $\endgroup$ Jan 2, 2019 at 14:28

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