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

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) ((10) !/6 !) (B) ((10) !/24) (C) ((9) !/24) (D) None of these

Tardigrade Admin · Accepted Answer

Selection of 6 guests = 10 C6 Permutation of 6 on round table =5 ! Permutation of 4 on round table =3 Then, total number of arrangements = 10 C6 ⋅ 5 ! ⋅ 3 !=((10) !/6 ! ⋅ 4 !) ⋅ 5 ! ⋅ 3 ! =((10) !/24)

Q. 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

$\frac{( 10 )!}{6 !}$

$\frac{( 10 )!}{24}$

$\frac{( 9 )!}{24}$

None of these

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
$=^{10} C_{6} \cdot 5! \cdot 3! = \frac{( 10 )!}{6 ! \cdot 4 !} \cdot 5! \cdot 3!$
$= \frac{( 10 )!}{24}$

Q. 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

6!(10)!​

24(10)!​

24(9)!​