There are 3 teachers and 6 students who will sit on the 9 available seats. many arrangements they sit if each teacher is flanked by 2 students


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Question Number 115769 by bemath last updated on 28/Sep/20

$T h e r e a r e 3 t e a c h e r s a n d 6 s t u d e n t s$ $w h o w i l l s i t o n t h e 9 a v a i l a b l e s e a t s . m a n y$ $a r r a n g e m e n t s t h e y s i t i f e a c h$ $t e a c h e r i s f l a n k e d b y 2 s t u d e n t s$
	Answered by mr W last updated on 28/Sep/20

	$m e t h o d 1 :$ $S ⧫ S ⧫ S ⧫ S ⧫ S ⧫ S$ $t h e t h r e e t e a c h e r s m a y t a k e 3 o f t h e$ $p l a c e s m a r k e d w i t h ⧫ . t h e r e a r e$ $C_{3}^{5} w a y s .$ $\Rightarrow C_{3}^{5} \times 3! \times 6! = 43200$ $m e t h o d 2 :$ $S_{1} {T S}_{2} {T S}_{3} {T S}_{4}$ $S_{1, 2, 3, 4} ⩾ 1$ $S_{1} + S_{2} + S_{3} + S_{4} = 6$ ${(x + x^{2} + x^{3} + . . .)}^{4} = \frac{x^{4}}{{(1 - x)}^{4}} = x^{4} \underset{k = 0}{\sum^{\infty}} C_{3}^{k + 3} x^{k}$ $c o e f . o f x^{6} i s w i t h k = 2$ $C_{3}^{2 + 3} = C_{3}^{5}$ $\Rightarrow C_{3}^{5} \times 3! \times 6! = 43200$
	Commented by bemath last updated on 29/Sep/20

	$g a v e k u d o s . s a n t u y s i r$
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