Question

In how many ways can $8$ persons be seated on two round tables of capacity $5$ & $3$.

Answer 1

Step 1: We have select $5$ out of $8$ members to be seated on big table

no of ways =${}^8{C}_5=\frac{8!}{5!3!}=\frac{8\times 7\times 6}{6}=56$

Step 2: we arrange $5$ selected members around big table

No of ways $=(5-1)!=4!$

Step 3: we arrange $3$ remaining members around small table

No of ways $=(3-1)!=2!$

\therefore Total no ways $=\frac{8}{5}\times 4!\times 2!$

$=56\times 24\times 2$

$=56\times 48$
$=2688$

$8$ person can be seated around $2$ different table in $2688$ ways