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Question Number 108818 by bobhans last updated on 19/Aug/20

$\frac{B o b h a n s}{Δ}$ $(1) L e t n b e a p o s i t i v e i n t e g e r, a n d l e t x a n d y b e p o s i t i v e r e a l$ $n u m b e r s u c h t h a t x^{n} + y^{n} = 1 . P r o v e t h a t$ $(\underset{k = 1}{\sum^{n}} \frac{1 + x^{2 k}}{1 + x^{4 k}}) (\underset{k = 1}{\sum^{n}} \frac{1 + y^{2 k}}{1 + y^{4 k}}) < \frac{1}{(1 - x) (1 - y)}$ $(2) A l l t h e l e t t e r s o f t h e w o r d^{'} E A M C O T^{'} a r e a r r a n g e d i n d i f f e r e n t$ $p o s s i b l e w a y s . T h e n u m b e r o f s u c h a r r a n g e m e n t i n w h i c h$ $n o t w o v o w e l s a r e a d j a c e n t t o e a c h o t h e r i s___$
	Answered by john santu last updated on 19/Aug/20

	$\frac{⊸ J S ⊸}{♢}$ $(2) w e n o t e t h a t t h e r e a r e 3 c o n s o n a n t a n d$ $3 v o w e l s A, E a n d O . S i n c e n o t w o v o w e l s$ $h a v e t o b e t o g e t h e r, t h e p o s s i b l e c h o i c e f o r$ $v o w e l s a r e t h e p l a c e s m a r k e d a s^{'} ♢^{'}$ $♢ M ♢ C ♢ T ♢, t h e s e v o w e l s c a n b e$ $a r r a n g e d i n P_{3}^{4} w a y s, 3 c o n s o n a n t c a n b e$ $a r r a n g e d i n 3! w a y s . H e n c e t h e r e q u i r e d$ $n u m b e r o f w a y s i s 3! \times P_{3}^{4}$ $= 6 \times \frac{4!}{(4 - 3)!} = 6 \times 24 = 144$
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