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02 Jul 2021, 06:15
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
57% (01:12) correct
43%
(01:54)
wrong
based on 663
sessions
History
Date
Time
Result
Not Attempted Yet
A man walks to his home from his current location on the rectangular grid shown. If he may choose to walk north or east at any corner, but may never move south or west, how many different paths can the man take to get home?
(A) 12
(B) 24
(C) 32
(D) 35
(E) 64
Attachment:
2021-06-17_23-29-08.png [ 19.97 KiB | Viewed 12133 times ]
02 Jul 2021, 06:29
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Bunuel
Allowed movements
North and East
CONCEPT:
Number of ways to reach from one point to other in a grid = \(^{(H+V)}C_H\) OR \(^{H+V}C_V\)
where,
V = Total number of Vertical Paths to reach from A to B
H = Total number of Horizontal Paths to reach from A to B
Number of ways to reach from one point to other in a grid = \(^{(H+V)}C_H\) OR \(^{H+V}C_V\)
where,
V = Total number of Vertical Paths to reach from A to B
H = Total number of Horizontal Paths to reach from A to B
Here, V = 3
and, H = 4
Number of ways to reach from one point to other in a grid = \(^{(4+3)}C_4\) OR \(^{(4+3)}C_3\) \(= \frac{7!}{3!*4!} = 35\)
Answer: Option D
Check Similar Official Question and explanation for this formula here
General Discussion
12 Jul 2021, 07:21
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The man has to move North 3 times and East 4 times for a total of 7 decisions.
These can be arranged in any sequence, that is 7! ways.
However, there is no difference between any of the North decisions inter-se and same for East decisions as well.
Hence, we have to divide the equation by 4! and 3!.
Answer = 7!/3!4! = 35 ways.
Answer - D
These can be arranged in any sequence, that is 7! ways.
However, there is no difference between any of the North decisions inter-se and same for East decisions as well.
Hence, we have to divide the equation by 4! and 3!.
Answer = 7!/3!4! = 35 ways.
Answer - D
