In how many ways can 3 men and 5 women be seated in a row if 3 specific women cannot sit next to each other? My solution: arrangements when the 3 women cannot sit next to eachother = 5! ×^6 P_3 = 14,400 Solution in book: arrangements with no restrictions = 8! arrangements with the 3 women seated next to eachother = 3! × 6! arrangements when the 3 women cannot sit next to eachother = 8! − (3! × 6!) = 36,000 I′m confused. Can I please get some explanation why my solution is wrong?


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Question Number 125191 by Don08q last updated on 08/Dec/20

$In how many ways can 3 men and 5$ $women be seated in a row if 3 specific$ $women c a n n o t sit next to each other ?$ $My solution :$ $arrangements when the 3 women$ $c a n n o t sit next to eachother$ $= 5! \times^{6} P_{3}$ $= 14, 400$ $Solution in book :$ $arrangements with no restrictions = 8!$ $arrangements with the 3 women$ $seated next to eachother = 3! \times 6!$ $arrangements when the 3 women$ $c a n n o t sit next to eachother$ $= 8! - (3! \times 6!)$ $= 36, 000$ $I^{'} m confused . Can I please get some$ $explanation why my solution is wrong ?$
Commented by liberty last updated on 09/Dec/20

$(∙) - W_{1} - W_{2} - W_{3} -$ $t r e a t 3 m e n a n d 2 r e m a i n i n g w o m e n a s$ $i d e n t i c a l^{'} x^{'} . S i n c e W_{1}, W_{2} a n d W_{3} n o$ $a d j a c e n t \Rightarrow - W_{1} \overset{x}{-} W_{2} \overset{x}{-} W_{3} -$ $t h e 3 r e m a i n i n g^{'} x^{'} s c a n b e p u t t h e 4 p l a c e s$ $a r b i t r a r e l y i n (\begin{matrix} 3 + 4 - 1 \\ 3 \end{matrix}) = (\begin{matrix} 6 \\ 3 \end{matrix}) = 20$ $t h e n u m b e r o f w a y a r r a n g e m e n t i s$ $g i v e n b y 20 \times 3! \times 5! = 120 \times 120 = 14400$
Commented by mr W last updated on 09/Dec/20

$b o t h o f y o u a r e w r o n g s i r s .$ $y o u t r e a t e d t h e c a s e s t h a t ‘ ‘ n o t w o o f$ $t h e t h r e e w o m e n a r e n e x t t o e a c h$ ${o t h e r}^{″} . b u t t h e q u e s t i o n a s k e s t h a t$ $‘ ‘ a l l t h e t h r e e w o m e n a r e n o t n e x t$ $t o e a c h {o t h e r}^{″} . t h a t m e a n s a c c o r d i n g$ $t o t h e q u e s t i o n w e s h o u l d o n l y e x c l u d e$ $t h e c a s e s t h a t a l l t h e t h r e e w o m e n$ $a r e t o g e t h e r . b u t i n y o u r s o l u t i o n s$ $y o u e x c l u d e a l s o t h e c a s e s t h a t t w o$ $o f t h e t h r e e w o m e n a r e t o g e t h e r .$ $t h e s o l u t i o n g i v e n i n b o o k i s c o r r e c t .$
Commented by mr W last updated on 09/Dec/20

$a g a i n : a c c o r d i n g t o q u e s t i o n i t i s$ $a l l o w e d w h e n t w o o f t h e t h r e e w o m e n$ $a r e t o g e t h e r . b u t t h i s i s n o t a l l o w e d$ $i n y o u r s o l u t i o n s .$
Commented by liberty last updated on 11/Dec/20

$y e a h . . . . i m i s u n d e r s t a n d i n g t h e q u e s t i o n .$
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