Question

A tea party is arranged for $16$ people along two sides of a large table with $8$ chairs on each side. Four men to sit on one particular side and two on the other side. In how many ways can they be seated?

• A. $\frac{8!8!10!}{(4!{)}^{2}6!}$
• B. $\frac{8!8!10!}{6!}$
• C. $\frac{8!8!10!}{4!!}$
• D. $\frac{8!8!10!}{4!6!}$

Answer 1

The correct option is D $\frac{8!8!10!}{4!6!}$

There are $8$ chairs on each side of the table. Let sides be represented by $A$ and $B$. Let four persons sit on side $A$, then number of ways arranging $4$ persons on $8$ chairs on side $A={}^8{P}_4$ and then two persons sit on side $B$, then the number of ways arranging $2$ persons on $8$ chairs on side $B={}^8{P}_2$ and arranging the remaining $10$ persons in remaining $10$ chairs in $10!$ ways.
Hence the total number of ways in which the persons can be arranged
$={}^8{P}_4\times {}^8{P}_2\times 10!$
$=\frac{8!8!10!}{4!6!}$