Question

Twelve friends go out for a dinner to $UTSAV$ restaurant. There are two circular tables one with $7$ chairs and one with $5$ chairs. In how many ways can the group settle down themselves for dinner?

• A. $\frac{12!}{7!\times 5!}$
• B. $\frac{12!}{35}$
• C. $12!$
• D. $12!5!7!$

Answer 1

Answer: (B)

$\frac{12!}{35}$

Number of ways of selecting $7$ people for first round table $={}^{12}{C}_5$

Number of ways of selecting remaining $5$ people for second round table $={}^5{C}_5$

Arranging $7$ people on first round table $=(7-1)!=6!$

Arranging $5$ people on second round table $=(5-1)!=4!$

Hence the answer is $={}^{12}{C}_5\times {}^5{C}_5\times 6!\times 4!$

$=\frac{12!\times 5!\times 6!\times 4!}{7!\times 5!\times 5!\times 0!}=\frac{12!}{7\times 5}=\frac{12!}{35}$

Hence the correct answer is $\frac{12!}{35}$.