Standing on the origin of an xy-coordinate plane, John takes a 1-unit step at random in one of the following 4 direction

[BTGmoderatorDC]

Standing on the origin of an xy-coordinate plane, John takes a 1-unit step at random in one of the following 4 directions: up, down, left, or right. If he takes 3 more steps under the same random conditions, what is the probability that he winds up at the origin again?

(A) 7/64
(B) 9/64
(C) 11/64
(D) 13/64
(E) 15/64


OA B

Source: Manhattan Prep

[swerve]

Standing on the origin of an xy-coordinate plane, John takes a 1-unit step at random in one of the following 4 directions: up, down, left, or right. If he takes 3 more steps under the same random conditions, what is the probability that he winds up at the origin again?

(A) 7/64
(B) 9/64
(C) 11/64
(D) 13/64
(E) 15/64


OA B

Source: Manhattan Prep

Try as follows:

Total no of cases\(=4^4\)
Way \(1- 1\) each of \(L, R, U, D -\) They can be arranged in \(4!\) ways
Way \(2- 2\) each of \(R \& L..RRLL -\) They can be arranged in \(\dfrac{4!}{2!\cdot 2!}= 6\)
Way \(2- 2\) each of \(D \& U..DDUU -\) They can be arranged in \(\dfrac{4!}{2!\cdot 2!}= 6\)

\(36\) ways possible / \(4\cdot 4 \cdot 4\cdot 4\)
\(=\dfrac{9}{64}\)

[Scott@TargetTestPrep]

Standing on the origin of an xy-coordinate plane, John takes a 1-unit step at random in one of the following 4 directions: up, down, left, or right. If he takes 3 more steps under the same random conditions, what is the probability that he winds up at the origin again?

(A) 7/64
(B) 9/64
(C) 11/64
(D) 13/64
(E) 15/64


OA B

Solution:

Since each stop has 4 choices, 4 steps have 4^4 = 256 possible paths. We now need to figure out how many of these paths end up at the origin again. For example, if John goes up, up, down and down, he will end up at the origin again. Let’s denote up as U, down as D, left as L, and right as R. Therefore, in the example above, his path can be denoted as UUDD. However, any of John’s path that is a rearrangement of these 4 letters will also allow him to end up at the origin again, e.g., UDUD and DUUD. Therefore, we can list all the “distinct” paths that allow him to wind up at the origin and also figure out the number of ways to rearrange the letters in that path (here, we consider two paths to be “distinct” if they have at least one letter that is not the same).

Path Number of arrangements
UUDD 4!/(2! x 2!) = 24/4 = 6
RRLL 4!/(2! x 2!) = 24/4 = 6
UDRL 4! = 24

Therefore, there are 6 + 6 + 24 = 36 paths in which John ends up at the origin again,and hence the probability of doing so is 36/256 = 9/64.

Answer: B