Question 6 : The question seems to be wrong (not sure)

I think the question should be 2*10 instead of 2*12.
Otherwise none of the options will match.
The recursion should be this.
t(n) = t(n-1) +t(n-2) + 2* c(n-1)
and, c(n) = c(n-2) + t(n-2)

According to you what should be the answer

If the question is 2*10 instead of 2*12, then the answer can be 1255 (option b).

The answer has not come. No one knows the correct answer

Obviously

Could you explain how you made the sequence?

Before starting with the explanation consider all the floors as 4*6 matrices.
The 1st and the 9th floors are obvious.
Now lets figure out the 2nd floor. There is one U(up staircase) in the 1st floor at (5,1) position. To come there there should be D(down staircase) in the 2nd floor in (5,1) , (5,2) or (5,3) position. There are only two such floors (cards 5 and 6). Also the 2nd floor should have only one D. This eliminates card 5. Thus card 6 represents the 2nd floor.

After this try proceeding backwards in a similar fashion. Find the 8th floor then 7th and so on.

Ok,but isnt that question 4? I wanted to know how you got the answer to question 6

Oh yeah sorry for that.

Let t(n) be the number of ways to tile a 2*n floor.
Let c(n) be the number of ways to tile a 2*n floor such that the rightmost box is left in the upper row.
You can add a vertical 2*1 tile in one way to t(n-1) to get t(n)
                            or,
You can add  2 horizontal 2*1 tiles in one way to t(n-2) to get t(n)
                            or,
You can add a 3 box tile in two ways to c(n-2) to get t(n).
So, t(n) = t(n-1)+t(n-2)+c(n-2)

Similarly to get c(n),
You can add a 3 box tile in one way to t(n-2) to get c(n)
or,
You can add a a 2*1 horizontal tile to c(n-2) to get c(n).
So, c(n)=t(n-2) + c(n-2).

Wow.
I never thought it would be so difficult

Hmm Im not sure I understand every step that youre doing but I liked the idea of using recursion, and will try something on my own. But in your method, you are assuming only the arrangements present in previous arrangements to also be in T(n), with additions only at the ends. Will there not be some new arrangements in T(n) among the previous blocks themselves which were not present in the earlier terms like T(n-1)? Im not sure if I phrased it right but hope you get what Im saying

Is q.7 answer 56 and q.8 answer 03

@Sai1704 ..... No, there won't be any new arrangements in terms like t(n-1). All the arrangements have been counted. Take small values for 'n' like 3, 4,5 etc. You will start thinking the the lines of recursion. This is also a type of sequential counting which i told you in the question related with tiling the floors.

Hmm alright, but I just want to take an example to show what I mean,
For 2x4, an arrangement is possible using 2 3-square tiles and 1 rectangular tile like so (If I number each tile on the grid as
1 2 3 4
5 6 7 8
-Two 3 square tiles at 156 and 487
-One rectangular tile at 23
Now such an arrangement is not possible using only 2x3 right? Im not sure if thats why you used a c(n) as well, so could you just confirm that?

Yes. If you don't use c(n) then such possibilities will be left out. But c(n) covers these type of cases.