Question
A person invites a party of 10 friends at dinner and place them so that 4 are on one round table and 6 on the other round table. The number of ways in which he can arrange the guests is
A.
$$\frac{{\left( {10} \right)!}}{{6!}}$$
B.
$$\frac{{\left( {10} \right)!}}{{24}}$$
C.
$$\frac{{\left( {9} \right)!}}{{24}}$$
D.
None of these
Answer :
$$\frac{{\left( {10} \right)!}}{{24}}$$
Solution :
Selection of 6 guests $$ = {\,^{10}}{C_6}$$
Permutation of 6 on round table $$= 5!$$
Permutation of 4 on round table $$= 3!$$
Then, total number of arrangements $$ = {\,^{10}}{C_5} \cdot 5! \cdot 3!$$
$$ = \frac{{\left( {10} \right)!}}{{6! \cdot 4!}} \cdot 5! \cdot 3! = \frac{{\left( {10} \right)!}}{{24}}.$$