Suppose 32 objects are placed along a circle at equal distances. Let N be the number of ways 3 objects can be chosen from them so that two of the three chosen objects are neither adjacent nor diametrically opposite. Then the product of the digits of sum of digits of N is equal to 

We can  choose 3 objects out of 32 objects in (32,3) ways. Among these possibilities 

 all the three would be together in 32 cases; 

exactly two will be together in 32 x 28 cases. 

Thus three objects can be chosen such that no two adjacent in (32,3) - 32 - (32 x 28) ways.,

 further, two objects will be diametrically opposite in 16 ways and the third object would be on either side of the diameter  in a non adjacent portion in 32 - 6 = 26 ways.

$$\begin{gathered} N={}^{32}C_3-(32+32\times 28+16\times 26) \\ =3616 \end{gathered}$$