DSE Permutation and Combination Problem?

Let R₁, R₂ be the radius of the first and the last circle respectively.

(a)
Case 1：
R₁：Odd number　　　　R₂：Even number
The distance is odd number

Case 2：
R₁：Even number　　　　R₂：Odd number
The distance is a odd number

The number of arrangement = 2 C(4,1) C(3,1) P(5,3) = 1440

(b)
Case 1：
R₁ = 1　　　　R₂ = 6
The other circles = C(4,3) = 4
(1,2,3,4,6　　　1,2,3,5,6　　　1,2,4,5,6　　　1,3,4,5,6)

Case 2
R₁ = 2　　　　R₂ = 7
The other circles = C(4,3) = 4
(2,3,4,5,7　　　2,3,4,6,7　　　2,3,5,6,7　　　2,4,5,6,7)

The probability = (4 + 4)/1440 = 1/180