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