There are 12 students in a party. Five of them are girls. In how many ways can these 12 students be arranged in a row if (i) there are no restrictions? (ii) the 5 girls must be together (forming a block)? (iii) no 2 girls are adjacent (iv) between two particular boys A and B , there no boys but exactly 3 girls?


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Question Number 125208 by liberty last updated on 09/Dec/20

$T h e r e a r e 12 s t u d e n t s i n a p a r t y . F i v e o f$ $t h e m a r e g i r l s . I n h o w m a n y w a y s c a n$ $t h e s e 12 s t u d e n t s b e a r r a n g e d i n a r o w i f$ $(i) t h e r e a r e n o r e s t r i c t i o n s ?$ $(i i) t h e 5 g i r l s m u s t b e t o g e t h e r (f o r m i n g a b l o c k) ?$ $(i i i) n o 2 g i r l s a r e a d j a c e n t$ $(i v) b e t w e e n t w o p a r t i c u l a r b o y s A a n d B$ $, t h e r e n o b o y s b u t e x a c t l y 3 g i r l s ?$
	Answered by john_santu last updated on 09/Dec/20

	$(i) 12!$ $(i i) 5! (7 + 1)! = 5! 8!$ $(i i i) \underset{1}{-} G_{1} \underset{2}{\overset{x}{-}} G_{2} \underset{3}{\overset{x}{-}} G_{3} \underset{4}{\overset{x}{-}} G_{4} \underset{5}{\overset{x}{-}} G_{5} \underset{6}{-}$ $= 5! 7! (\begin{matrix} 6 + 3 - 1 \\ 3 \end{matrix}) = 5! 7! (\begin{matrix} 8 \\ 3 \end{matrix})$ $(i v) A G G G B - - - - - - -$ $- A G G G B - - - - - -$ $- - A G G G B - - - - -$ $- - - A G G G B - - - -$ $- - - - A G G G B - - -$ $- - - - - A G G G B - -$ $- - - - - - A G G G B -$ $- - - - - - - A G G G B$ $t h e n u m b e r o f w a y s i s$ $= 2 \times 8 \times C_{3}^{5} \times 3! \times 7!$ $= 12 \times 8! \times 10 = 120 \times 8!$
	Commented by liberty last updated on 09/Dec/20

	$y e s . . c o r r e c t . . t h a n k s$
	Commented by malwan last updated on 16/Jan/21

	$p l e a s e$ $I {c a n}^{'} t u n d e r s t a n d n u m b e r (i i i)$ $(\begin{matrix} 6 + 3 - 1 \\ 3 \end{matrix}) w h a t i s i t ? 6 ? 3 ?$ $i f g i r l s = 3 a n d b o y s = 6 t h e n$ $w h a t i s t h e s o l u t i o n ?$ $a n d w h a t i f g i r l s = n a n d$ $b o y s = m; m > n ?$
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