Question:
A business man hosts a dinner to 21 guests. He is having 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests?
Solution:
A businessman hosts a dinner for 21 guests.
15 people can be accommodated at one table in ${ }^{21} \mathrm{C}_{15}$ ways. They can arrange themselves in $(15-1) !=14 !$ ways.
The remaining 6 people can be accommodated at another table in ${ }^{6} \mathrm{C}_{6}$ ways. They can arrange themselves in $(6-1) !=5 !$ ways.
$\therefore$ Total number of ways $={ }^{21} C_{15} \times{ }^{6} C_{6} \times 14 ! \times 5 !={ }^{21} C_{15} \times 14 ! \times 5 !$