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Question Number 124806 by bramlexs22 last updated on 06/Dec/20

	Answered by mr W last updated on 06/Dec/20

	$(i)$ $6 = 1 + 5 = 2 + 4 = 3 + 3$ $\Rightarrow C_{1}^{6} \times 4! + C_{2}^{6} \times 3! + C_{3}^{6} / 2 \times 2! \times 2! = 274$ $(i i)$ $6 = 1 + 1 + 4 = 1 + 2 + 3 = 2 + 2 + 2$ $\Rightarrow C_{1}^{6} C_{1}^{5} / 2 \times 3! + C_{1}^{6} C_{2}^{5} \times 2! + C_{2}^{6} C_{2}^{4} / 3! = 225$
	Commented by john_santu last updated on 06/Dec/20

	$t h e t a b l e s a r e i n d i s t i n g u i s h a b l e s i r$
	Commented by mr W last updated on 06/Dec/20

	$y e s, n o w i s e e i t . p l e a s e c h e c k a g a i n .$
	Commented by john_santu last updated on 06/Dec/20

	$(i i) i t h i n k = \frac{6 \times 5 \times 6}{2} + 6 \times 10 \times 2 + \frac{15 \times 6}{6}$ $= 90 + 120 + 15 = 225$
	Commented by john_santu last updated on 06/Dec/20

	$o k s i r . t h a n k y o u$
	Commented by bramlexs22 last updated on 06/Dec/20

	$t h a n k y o u b o t h s i r$
	Answered by john_santu last updated on 06/Dec/20

	$(i) F o r 2 t a b l e s t h e r e a r e 3 c a s e s t o c o n s i d e r$ $a c c o r d i n g t o t h e n u m b e r s o f p e o p l e t o b e$ $s e a t e d a r r o u n d t h e 2 r e s p e c t i v e t a b l e s, n a m e l y$ $(1) 5 + 1 (2) 4 + 2 (3) 3 + 3$ $c a s e (1) T h e r e C_{5}^{6} \times 4! \times 0! = 144$ $c a s e (2) T h e r e C_{4}^{6} \times 3! \times 1! = 90$ $c a s e (3) T h e r e \frac{C_{3}^{6}}{2!} \times 2! \times 2! = 40$ $t h e d e s i r e d n u m b e r o f a r r a n g e m e n t s$ $i s 144 + 90 + 40 = 274 .$
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