Question

A tea party is arranged for $16$ people along two sides of a large table with $8$ chair on each side. Four men want to sit on one particular side and two on the other side. The number of ways in which they can be seated is

• A. $\frac{6!8!10!}{4!6!}$
• B. $\frac{8!8!10!}{4!6!}$
• C. $\frac{8!8!6!}{6!4!}$
• D. None of these

Answer 1

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

There are $8$ chair on each side of the table.

Let the sides be represented by $A$ and $B$.

Let four persons sit on side $A$, then number of ways of arranging 4 persons on 8 chairs on side $A={}^8{P}_4$

And two persons sit on side $B$.

The number of ways of arranging $2$ persons on $8$ chairs on side $B={}^8{P}_2$

The remaining $10$ persons can be arranged in remaining $10$ chairs in $10!$ ways.
Hence, the total number of ways in which the persons can be arranged is
${}^8{P}_4\times {}^8{P}_2\times 10!=\frac{8!8!10!}{4!6!}$