Question

Five boys and five girls form a line. Match the number of ways of making the seating arrangement given in Column II under the conditions given in Column I.

Column-I		Column-II
(i)	Boys and girls sits alternate.	(p)	$5!\times 6!$
(ii)	No two girls sit together.	(q)	$10!-5!6!$
(iii)	All the girls are never together.	(r)	${\left(5!\right)}^2+{\left(5!\right)}^2$

• A. (i) $\to$ (q), (ii) $\to$(r), (iii)$\to$ (p)
• B. (i) $\to$ (r), (ii) $\to$(q), (iii)$\to$ (p)
• C. (i) $\to$ (p), (ii) $\to$(q), (iii)$\to$ (r)
• D. (i) $\to$ (r), (ii) $\to$(p), (iii)$\to$ (q)

Answer 1

Answer: (D)

(i) Total arrangements when boys and girls sit at alternate positions
$=(5!{)}^2+(5!{)}^2$
(ii) Seating arrangement should be

$\Rightarrow$ There are six positions of seating for girls.
$\therefore$ Required number of ways $={}^5{P}_5\times {}^6{P}_5=5!\times 6!$
(iii) Required no. of ways = Total arrangements - (No. of ways when all the girls sit together) $=10!-5!6!$