Eight eligible bachelor and seven beautiful models happen randomly to have purchased single seats in the same 15−seats row of theather. On the average , how many pairs of adjacent seats are ticketed for marriageable couples ?


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Question Number 131459 by bramlexs22 last updated on 05/Feb/21

$E i g h t e l i g i b l e b a c h e l o r a n d s e v e n b e a u t i f u l$ $m o d e l s h a p p e n r a n d o m l y t o h a v e$ $p u r c h a s e d s i n g l e s e a t s i n t h e s a m e 15 - s e a t s$ $r o w o f t h e a t h e r . O n t h e a v e r a g e, h o w m a n y$ $p a i r s o f a d j a c e n t s e a t s a r e t i c k e t e d$ $f o r m a r r i a g e a b l e c o u p l e s ?$
	Answered by bemath last updated on 05/Feb/21

	$m o r e g e n e r a l l y w i t h b e l e m e n t s$ $o f o n e k i n d a n d m o f a n o t h e r$ $r a n d o m l y a r r a n g e d i n a l i n e, t h e$ $e x p e c t e d n u m b e r o f u n l i k e$ $a d j a c e n t s e l e m e n t s i s$ $(m + b - 1) [\frac{b m}{(m + b) (m + b - 1)} + \frac{m b}{(m + b) (m + b - 1)}] =$ $\frac{2 m b}{m + b} . I n o u r q u e s t i o n b = 8, m = 7$ $g i v i n g \frac{2.8.7}{15} = 7 \frac{7}{15}$
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