Question

A number of 18 guests have to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on the other side. let the number of ways in which the sitting arrangements can be made is equal to ${}^{11}{C}_k\times m!\times m!$ arrangements. Find m+k ?

Answer 1

Out of 18 guests, 9 are to be seated on side A and rest 9 on side B.
Now
out of 18 guests, 4 particular guests desire to sit on one particular
side, say side A, and other 3 on other side B. Out of rest $18-4-3=11$
guests, we can select 5 more for side A and rest 6 can be seated on
side B. Selection of 5 out of 11 can be done in ${}^{11}{C}_5$ ways. Nine
guests on each side of table can be seated in $9!\times 9!$ ways.
Thus, there are total ${}^{11}{C}_5\times 9!\times 9!$ arrangements.