Question
A teaparty is arranged for 16 people along two sides of a large table with 8 chairs 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 :
$$\frac{{8!8!10!}}{{4!6!}}$$
Solution :
There are 8 chairs 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 then two persons sit on side $$B.$$ The number of ways of arranging 2 persons on 8 chairs on side $$B = {\,^8}{P_2}$$ and 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
$$ = {\,^8}{P_4} \times {\,^8}{P_2} \times 10! = \frac{{8!8!10!}}{{4!6!}}$$