There are 10 monkeys. The total number of ways in which these monkeys can be seated around a round table so that exactly one monkey sits between 2 particular monkeys, is equal to

Question

There are 10 monkeys. The total number of ways in which these monkeys can be seated around a round table so that exactly one monkey sits between 2 particular monkeys, is equal to (A) 8 ! (B)  2 ! ⋅ 8 ! (C) 3 ! ⋅ 7 ! (D) 2 ! ⋅ 7 !

Tardigrade Admin · Accepted Answer

Let 2 monkeys be M 1 M 2 M 1- M 2 arrow 8 C 1 (one can sit between them out of remaining 8) Put - M 1- M 2 ldots ldots ldots ldots ldots Arrangement about round table = 8 C 1 × 7 ! × 2 ! (2 != text arrangement of M 1 M 2)

Q. There are 10 monkeys. The total number of ways in which these monkeys can be seated around a round table so that exactly one monkey sits between 2 particular monkeys, is equal to

$8!$

$2! \cdot 8!$

$3! \cdot 7$ !

$2! \cdot 7$ !

Let 2 monkeys be $M_{1} M_{2}$
$M_{1} - M_{2} \to^{8} C_{1}$ (one can sit between them out of remaining 8)
Put - $M_{1} - M_{2} \dots\dots\dots\dots\dots$
Arrangement about round table
$=^{8} C_{1} \times 7! \times 2! (2! = arrangement of M_{1} & M_{2})$

Q. There are 10 monkeys. The total number of ways in which these monkeys can be seated around a round table so that exactly one monkey sits between 2 particular monkeys, is equal to

8!

2!⋅8!

3!⋅7 !

2!⋅7 !

Let 2 monkeys be M1​M2​ M1​−M2​→8C1​ (one can sit between them out of remaining 8) Put - M1​−M2​…………… Arrangement about round table =8C1​×7!×2!(2!= arrangement of M1​&M2​)

$8!$

$2! \cdot 8!$

$3! \cdot 7$ !

$2! \cdot 7$ !

Let 2 monkeys be $M_{1} M_{2}$
$M_{1} - M_{2} \to^{8} C_{1}$ (one can sit between them out of remaining 8)
Put - $M_{1} - M_{2} \dots\dots\dots\dots\dots$
Arrangement about round table
$=^{8} C_{1} \times 7! \times 2! (2! = arrangement of M_{1} & M_{2})$