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I too really confused about the problem ?? can u pls explain ?

Thanks in advance

Explain of (3, 1, 2):

There are five kind of mountains for n = 3. 1st mountain has 3 peek of height 1, 0 peek of height 2, 0 peek of height 3.
2nd and 3rd mountain has 1 peek of height 1, 1 peek of height 2, 0 peek of height 3.
4th mountain has 0 peek of height 1, 2 peek of height 2, 0 peek of height 3.
5th mountain has 0 peek of height 1, 0 peek of height 2, 1 peek of height 3.

So the number of mountains that has exactly r = 1 peek of height k = 2 is 2 (2nd and 3rd).

Other examples of n = 3.
(3, 0, 2) = 2 (1st and 5th)
(3. 0, 1) = 2 (4th and 5th)
(3, 2, 1) = 0 (none)
(3, 1, 3) = 1 (5th)
(3, 0, 4) = 5 (all)
(3, 1, 4) = 0 (none)

I hope this helps.
----
Sory for my poor English.

Thanks buddy.... Now i can give a try

Is there any DP solution or anything......

Yes, this is a dp problem.

I basically optimized dp[x][y][p][a]..
..ie how many ways I can come to coordinate x,y with p peaks and 'a' indicates last move(down hill or uphill ).

So ultimately the required answer is dp[2*N][0][R][0] // 0 -> downhill

some random test cases for debugging:
input : output :