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Case 1 3 Spaces L x x x L the 3 OCA can be arranged 3! = 6

Case 2

2 spaces L x x L x or x L x x L can 3C2 * 3C2 ways = 9

Case 3

1 Space will be 3 ways

total number of ways will be 6 * 9 * 3 = 162 ways

RC can u please post the OA

rxs, I see you approached this problem the long way with favorable outcome approach. I agree with case 1 but for cases 2 and 3, there is a mistake in your reasoning

case2 LxxLx
xLxxL
For each of the above 2, you have 3! ways of arranging 3 other letters. So you have 3!*2 = 12

case3 LxLxx
xLxLx
xxLxL
Once again, for each of these possible ways of positioning "L", you have 3! ways of doing so. So it is 3!*3 = 18

So you have total favorable outcomes = case(1+2+3) = 6+12+18 = 36

Case 1 3 Spaces L x x x L the 3 OCA can be arranged 3! = 6

Case 2

2 spaces L x x L x or x L x x L can 3C2 * 3C2 ways = 9

Case 3

1 Space will be 3 ways

total number of ways will be 6 * 9 * 3 = 162 ways

RC can u please post the OA

rxs, case 1 is good. But there is a problem in your case 2 and case 3.

For case 2:
For L _ _ L _ : You will have 3! ways, not 3C2
LOCLA;
LCOLA
LAOLC
LOALC
LACLO
LCALO

Similarly, for _ L _ _ L : You will have another 3! ways.

Case 3:
It should not be just 3 ways.
You will have:
L _ L _ _ : 3! ways for this arrangement
_ L _ L _ : 3! ways for this arrangement
_ _ L _ L : 3! ways for this arrangement

In the end you need to add all the ways not multiply.
So you get 6(3!) = 6*6 = 36