We can use numbers such as 123654, the aim to make sure on the right of 6,5,4, there are only 0, 1 or 2 digits smaller than them. The important numbers to place are 4, 5, and 6, since 3 2 1 can be placed anywhere. We will accept _ 4 5 6 _ _ for example but not _ 4 6 5 _ _ as on the right of 6 there are 3 numbers smaller than it.

I split the cases according to where the placements of 4, 5, and 6 are.

Case 1: _ _ _ 4 5 6. We can scramble 4 5 and 6. Thus there are 3*2 cases for the first 3 blanks and 3*2 cases for the last 3 digits. 6*6= 36 cases.

Case 2: _ 4 5 6 _ _. Note we are allowed to move the 5 towards the right and 6 towards the right. The last 4 placements can be randomized except for the _465_ _, _46_5_, and _46_ _5 case. Then we have 4*3 - 3 = 9 scenarios of 6 cases each, 54 cases.

Case 3: _ _ 4 5 6 _, or _ _ 546 _, or _ _ 564 _. In all cases, we are allowed to randomize the last 3 placements (so the last two scenarios overlap, only counts as one). Thus this is 2 * 6 * 6 = 72 cases.

In total 36 + 54 + 72 = 162.

Ans: D
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