Question

No. of ways of seating 20 persons at two round tables, seating 10 to each table is

• A. $\frac{20!}{100}$
• B. $\frac{20!}{81}$
• C. $\frac{20!}{10}$
• D. $\frac{20!}{9}$

Answer 1

The correct option is A $\frac{20!}{100}$

We know that,

$n$ objects can be arranged around a circle in $(n-1)!$

If a If arranging these $n$ objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number
i.e., number of arrangements.$=\frac{(n-1)!}{2}$

But I can choose the 10 people to sit in the first table in $\begin{align}

${}^{20}{C}_{10!}=\frac{20!}{10!(20-10)!}$

$=\frac{20!}{10!\times 10!}$

After selecting 10 people of two times can be made to sit the first and second table is

$=(10-1)!\times (10-1)!$

$=9!\times 9!$

Hence, the total number of ways

$=\frac{20!}{10!\times 10!}\times 9!\times 9!$

$=\frac{20!}{10\times 9!\times 10\times 9!}\times 9!\times 9!$

$=\frac{20!}{10\times 10}$

$=\frac{20!}{100}$

Hence, this is the answer..