Problem 6

There are ${12 \choose 5}\cdot{7 \choose 4}\cdot{3\choose 3} = 27720 = {12\choose 4}\cdot{8\choose 5}\cdot{3\choose 3} = {12\choose 3}\cdot{9\choose4}\cdot{5\choose 5}...$ possibilities to seat 12 people assuming that the order of seating is inmaterial.

Note.- Realize that the numerators in combinations of above are $12!$ and the denominators are $5!\cdot4!\cdot3!$

No. Of ways of selecting 3 people for the smallest table out of 12 people is $^{12}C_3$ . Now, to select 4 people from the remaining for the second smallest table we take $^{12-3}C_4$ . Then atlast the remaining 5 people gets alloted to the larget table by $^{12-3-4}C_5$ .
So, the final answer is $^{12}C_3 × ^{9}C_4 × ^5C_5 = 220 × 126 × 1 = \color{#3D99F6}{\boxed{27720}}$ .