Question

There are $6$ boys and $5$ girls. There are two round tables with $7$ chairs and $4$ chairs.

• A. Number of ways of arranging all of them is $\frac{11!}{28}$
• B. Number of ways of arranging all of them so that all girls are at same table is $\frac{(6!{)}^2}{8}$
• C. Number of ways of arranging all of them so that all boys are at same table is $\frac{5!6!}{4}$
• D. Number of ways of arranging all of them so that all boys are at one table or all girls are at one table is $6!\left(120\right)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Number of ways of arranging all of them is $\frac{11!}{28}$
B Number of ways of arranging all of them so that all girls are at same table is $\frac{(6!{)}^2}{8}$
C Number of ways of arranging all of them so that all boys are at same table is $\frac{5!6!}{4}$
D Number of ways of arranging all of them so that all boys are at one table or all girls are at one table is $6!\left(120\right)$

$\to$ Number of ways of arranging all is
$[{}^{11}{C}_7\times (7-1)!]\times [{}^4{C}_4\times (4-1)!]$
$=\frac{11!}{7!4!}\times 6!\times 3!$
$=\frac{11!}{28}$

$\to$ Number of ways of arranging all of them so that all girls are at same table is
$[{}^6{C}_2\times (7-1)!]\times \left[\right(4-1)!]$
$=\frac{(6!{)}^2}{8}$

$\to$ Number of ways of arranging all of them so that all boys are at same table is
$[{}^5{C}_1\times (7-1)!]\times \left[\right(4-1)!]$
$=\frac{5!6!}{4}$

$\to$ Number of ways of arranging all of them so that all boys are at one table or all girls are at one table is
$=\frac{(6!{)}^2}{8}+\frac{5!6!}{4}$
$=6!\left(120\right)$