This is a case of “Circular Permutation” where you cannot differentiate between clockwise and anticlockwise directions.

No. of arrangements = $\frac{1}{2}(18-1)!=\frac{1}{2}(17!)$

Hence option (d)

This is a type of “Different to Different” question.

5 prizes can be distributed among 4 students in $4^5$ ways.

Hence option (a)

If 1 is the smallest element of the set, all or none of the other elements can be selected in $2^{11}$ ways, as for each number from 2 to 12, there are two options, of getting selected or not getting selected. There are 11 such numbers 2-12 including both, therefore 2.2.2…..11 times = $2^{11}$

 Similarly, If 2 is the smallest element in the set, 4,6,8,10 and 12 can be selected in $2^5$ ways

 If 3 is the smallest element in the set 6,9,12 $2^3$ different sets

Required Solution = $2^{11}$ + $2^5$ + $2^3$ + $2^2$ + $2^1$ + $2^1$ + 6 = 2102

5 people can sit together around a round table in $(5-1)!=4!$ Ways. But in anticlockwise and clockwise arrangements each of them will have the same neighbour. So, no. of arrangements possible = $\frac{1}{2}(4!)=12$.

So, they dine together for 12 days.

Hence option (a)

Soln:

There are a total of 8 balls with 3 of one kind, 2 of other kind and rest 3 of the same kind. These can be arranged in $\frac{8!}{(3!\times 2!\times 3!)}=560$.