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+4 votes

**Direction for the question given below**

The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town

Neelam resides her bicycle from her house at A to her office at B, taking the shortest path, then the number of possible shortest paths that she can choose is

- 60
- 75
- 45
- 90
- 72

+2 votes

90

0 votes

The answer provided in the previous post is right but the explanation is incorrect.

The shortest path will be A → E → F→ B. Because the entire map is symmetric but the diagonal creates a new shorter path from E→ F. Diagonal is shorter than the sum of its perpendicular and base (Pythagoras Theorem’s Corollary).

Now to go from A to E, we have to travel right two times and travel downwards two times. So, we could for example, take R, D, R and D or we could take R, R, D, D. So how many such arrangements are possible ?? Lets list them down :-

R R D D, D D R R, R D R D, D R D R, R D D R, D R R D – Six arrangements possible

So the questions is **how many permutations/arrangements** of these four roads (R R D D) is possible given two of them are **repeating twice.**

It is : $\large \frac{^{4} P_{4}}{2!*2!} = \frac{4!}{2! * 2!} = 6$

Similarly, for F to B, we have to take four rights and two downs (R R R R D D). So the number of arrangements of these **six roads**, one which repeats **twice** and the other repeats **four times** is :

$\large \frac{^{6} P_{6}}{4!*2!} = \frac{6!}{4! * 2!} = 15$

Therefore, the total number of ways are : $\large 6 *15 = 90$.

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