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Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at $(i,j)$ then it can move to either $(i + 1, j)$ or $(i,j + 1)$.

How many distinct paths are there for the robot to reach the point $(10,10)$ starting from the initial position $(0,0)$?

  1. $^{20}\mathrm{C}_{10}$
  2. $2^{20}$
  3. $2^{10}$
  4. None of the above
in Combinatory edited by
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4 Comments

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why option (b) 2^20 cannot be the answer?
longest distance 20 units a robot has to travel, with 2 choices at every step, i.e. up or right, so i believe 2^20 could be a possible answer. Please comment/explain.
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8 Answers

3 votes
3 votes
The answer for 84 is A. It is same as selecting 10 out of 20 elements.

2 Comments

20C10 or 210?

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It is 20 C 10.
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2 votes
2 votes

problems like these should be solved using movements ( UP , RIGHT )

see this 

1 vote
1 vote
To reach from $(0,0)$ to $(10,10)$ we have to take 20 steps.

Lets assume these 20 steps to be 20 places.

( _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ )

Out of these 20 places to have to choose 10 places for upper steps, the rest of the 10 places have to be for right steps.)

like one example is  ( _ u u _ _ u _ _ u _ _uuu_ _u_u_ _ _ _ u), the $‘– ‘$ places should be for right steps.

so, Option $A$  $2nCn$​ is the answer.
edited by

2 Comments

Good explanation. Mentioning that : 

Out of these 20 places to have to choose 10 places for upper steps, the rest of the 10 places have to be for right steps.)
 

Here $20 = 2*10$ i.e, $(2n)$ and choosing $10 $ i.e, $(n)$ and $10$ i.e, $(n) = (2n!)/(n!)(n!) = 2nCn$

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0 votes
0 votes
In order to move from (0,0) to (10,10)there will be total 20 movement out of which there will be 10 right,10 upward movement combination

so 20!/10!.10!= c(20,10)
Answer:

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