21 friends were invited for a party. Two round tables can accommodate 12 and 9 friends each. The number of ways of the seating arrangements of friends is .....

The arrangement involves partitioning the 21 friends into groups of 12 and 9, then arranging them at round tables where only relative order matters.

The number of ways to select 12 friends for the first table is $\binom{21}{12} = \frac{21!}{12!9!}$, with the remaining 9 assigned to the second table.

For each table with $n$ persons, circular permutations yield $(n-1)!$ distinct arrangements: $11!$ for the first table and $8!$ for the second.

Multiplying the selection and arrangement counts gives $\frac{21!}{12!9!} \cdot 11! \cdot 8!$.

Simplifying, since $12! = 12 \times 11!$ and $9! = 9 \times 8!$, we substitute to get $\frac{21!}{(12 \cdot 11!)(9 \cdot 8!)} \cdot 11! \cdot 8! = \frac{21!}{12 \cdot 9} = \frac{21!}{108}$.

Reducing the fraction $\frac{21!}{108}$ by the greatest common divisor (12) gives $\frac{35 \cdot 19!}{9}$, matching option C.