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In how many ways can one go from (−2,−2) to (2,2) without crossing (0,0), if in one step one can move either one co-ordinate horizontally or vertically ? (One only moves up and to the right and cannot come back.)

Here is my attempt:

The total number of ways one can go from (-2,-2) to (2,2) is 8C4 ways (or 70) as there are 4 right steps and 4 up steps. The number of ways (0,0) can be crossed starting from (-2,-2) and moving only right and up is
RRUU,UURR, URUR, RURU or 4 ways. So, total number of ways without crossing 70-4=66. I know I am wrong and seeking help.

Splendid Digital Solutions
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1 Answers1

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Note: based on your question I am assuming that by horizontal and vertical movements you are only referring to the right $R$ and upward $U$ movements.

You have found the total number of ways to go from $A(-2,-2)$ to $B(2,2)$ as $\binom{8}{4}$, which is correct. Now calculate the number of ways of going from $A$ to $O(0,0)$, this will be $\binom{4}{2}$ (because we will have to do $4$ steps out of which $2$ will be $U$). Likewise we will have $\binom{4}{2}$ ways to go from $O$ to $B$. Thus there are $\binom{4}{2}\binom{4}{2}$ number of paths that will include $O$. So you subtract this from the total.

Anurag A
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  • That is great. However, could you help me know what is faulty in the 4 ways I have earlier mentioned going from (-2,-2) to (2,2). – Splendid Digital Solutions Aug 08 '21 at 06:21
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    @SplendidDigitalSolutions firstly you have to note that going from $A$ to $O$ can be done in $6$ ways $RRUU, RURU, URUR, RUUR, RUUR, UURR$. Now going from $O$ to $B$ can be accomplished in any of these $6$ ways also. So in your counting you have missed the two cases and also not accounted for the fact that a journey from $A$ to $O$ can be combined with any journey from $O$ to $B$, e.g. $\color{blue}{RRUU}\color{red}{RRUU}$ or $\color{blue}{RRUU}\color{red}{UURR}$ – Anurag A Aug 08 '21 at 06:26