Gatebook

Question

I am not able to understand the last component in square brackets we have to minus from all combinations the combinations with that dotted line but intersection part i didn't get

Comments

I think in place of $14C7$ it should be $9C4$ in intersection part of option B

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I feel that last part is 9C5 * 4C3 *5C3  or 9C4 * 4C1 * 5C2

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The invalid paths are:
Go to $(5, 6)$, move to $(6, 6)$ and move to $(11, 11)$
Go to $(8, 8)$, move to $(8, 9)$ and move to $(11, 11)$
If you subtract these two from the total paths $\binom{20}{10}$, you'll be subtracting paths that traverse both the blocked lines 2 times. So you add all such paths to the final answer. (third term in the options)
So the answer should be $\binom{20}{10} - \binom{9}{4}*\binom{10}{5}-\binom{14}{7}*\binom{5}{3}+\binom{9}{4}*\binom{4}{2}*\binom{5}{3}$

This is same as option C if you change $\binom{14}{7}$ to $\binom{9}{4}$ in the third term.

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How you are getting 4C2??

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@Niharika1 As I said, the third term adds the paths that travel from first blocked line to the blocked line and then to the upper most corner. So for this, we first travel to the point $(5, 6)$ (starts from $(1, 1)$), then we move to $(6, 6)$. $\binom{4}{2}$ is the number of ways to move to $(8, 8)$ from $(6, 6)$.

Answer 1

yes option should be c replacing 14C7 part in intersection part to get 9C4 I think solution is considering 10 rows and  columns instead of 11 rows and 11  columns correct if  I am wrong