The number of ways in which 8 different flowers can be strung to form a garland so that 4 particular flowers are never separated is


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Question Number 11889 by murtaza8 last updated on 04/Apr/17

$The number of ways in which 8$ $different flowers can be strung to$ $form a garland so that 4 particular$ $flowers are never separated is$
	Answered by ajfour last updated on 04/Apr/17

	$\frac{(5 - 1)!}{2} (4!)$
	Commented by ajfour last updated on 04/Apr/17

	Commented by mrW1 last updated on 04/Apr/17

	$C a n y o u p l e a s e e x p l a i n y o u r s o l u t i o n ?$ $I t h i n k h e m e a n t t h a t 4 s p e c i a l f l o w e r s$ $s h o u l d a l w a y s c h o s e n t o g e t h e r .$
	Commented by mrW1 last updated on 04/Apr/17

	$i f w e t r e a t t h e s e 4 f l o w e r s a s o n e s u p e r f l o w e r .$ $5 f l o w e r s f o r m t h e g a r l a n d, t h e r e a r e$ $5! w a y . T o r e p l a c e t h i s s u p e r f l o w e r$ $w i t h o u r 4 a l w a y s - t o g e t h e r - f l o w e r s,$ $t h e r e a r e 4 w a y s . S o t h e s o l u t i o n i s$ $5! 4! = 2880$
	Commented by mrW1 last updated on 04/Apr/17

	$S o r r y, I u n d e r s t a n d t h a t t h e g a r l a n d$ $s h o u l d c o n s i s t o f 4 f l o w e r s .$
	Commented by ajfour last updated on 04/Apr/17

	$b u t t h o s e f o u r c a n b e a r r a n g e d$ $a m o n g t h e m s e l v e s . .$ $f o r 4 + (1 g r o u p) = 5 f l o w e r s$ $h a v e a c i c u l a r p e r m u t a t i o n o f$ $\frac{(5 - 1)!}{2} . A n d t h e f o u r n o t t o b e$ $s e p a r a t e d s h o u l d, i b e l i e v e, h a v e a$ $p e r m u t a t i o n o f 4! .$
	Commented by ajfour last updated on 04/Apr/17

	$i f t h e r e a r e 3 d i f f e r e n t f l o w e r s$ $t o f o r m a g a r l a n d w e h a v e$ $\frac{(3 - 1)!}{2} = 1 w a y a n d {t h a t}^{'} s t r u e .$
	Commented by prakash jain last updated on 04/Apr/17

	$If we need to arrange n object in a$ $circular permutation than number$ $of arragements are (n - 1)! like$ $seating on a round table .$ $ABC, CAB, BCA are same but$ $ACB is different from ABC .$ $For arrangements that can turned$ $upside down like a garland$ $so clockwise and anticlockwise$ $are equivalent,$ $number of permutation are \frac{(n - 1)!}{2}$
	Commented by mrW1 last updated on 06/Apr/17

	$t h a n k s!$
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