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It took me about 3.5 mins to solve...I forgot some cases initially

Total no of cases=4^4
Way 1- 1 each of L,R,U,D- They can be arranged in 4! ways..4!
Way 2- 2 each of R & L..RRLL- They can be arranged in 4!/2!*2!= 6
Way 2- 2 each of D & U..DDUU- They can be arranged in 4!/2!*2!= 6

36 ways possible/4*4*4*4
=9/64

I think it helps to think directions as numbers with opposite signs...In this case the sum of the 4 numbers should be 0
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Total ways to fill 4 steps with 4 directions ( Order matters) = 4P1 x 4P1 x 4P1 x 4P1= 4x4x4x4

Now Favourable Ways in which we can reach the origin after 4 steps starting from the origin= Scenario 1 + Scenario 2

Scenario 1) We take a different direction in each step - no direction is repeated = 4P1 x 3P1 x 2P1 x 1P1 =4x3x2x1

Scenario 2) We take only two directions in 4 steps such as we return to the origin, which means we can use any one direction only twice. Two sets of directions can fill up these four steps
a) Only R and L ( 2 steps max with R and 2 Steps max with L) = 4P2 MINUS when one direction repeats in more than 2

RRLL
RLRL
RLLR
LLRR
LRLR
LRRL

6 steps

b) Only T and D ( 2 steps max with T and 2 Steps max with D) = 6 Steps following the above approach

Thus a+b = 12

Scenario 1 + Scenario 2 = 24+12= 36

P ( Favourable/ Total) = 36/256= 9/64
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