Question

Seven person $P_1,P_2,\dots \dots ,P_7$ initially seated at chairs $C_1,C_2,\dots \dots ,C_7$ respectively. They all left there chairs symultaneously for hand wash. Now in how many ways they can again take seats such that no one sits on his own seat and $P_1$ sits on $C_2$ and $P_2$ sits on $C_3$ ?

• A. $60$
• B. $53$
• C. $54$
• D. $55$

Answer 1

Answer: (A)

$60$

Seven person $P_1,P_2,\dots \dots ,P_7$ initially seated at chairs $C_1,C_2,\dots \dots ,C_7$ respectively.

no one sits on his own seat and $P_1$ sits on $C_2$ and $P_2$ sits on $C_3$

$P_{1}and{P}_2$ has fixed seats and $C_{2}and{C}_3$ cannot sit in there own places

So,let fix persons in there respective sits and subtracting from total number

$5!-3\cdot 4!$(fixing one at a time)$+3\cdot 3!$(fixing 2 at a time and it is repeating before so we will add)$-3\cdot 2!$(fixing 3 at a time and it is repeating in fixing 2 so we will subtract)

$5!-3\cdot 4+3\cdot 3!-3\cdot 2!=60$