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Question Number 111690 by bemath last updated on 04/Sep/20

	Answered by john santu last updated on 04/Sep/20
	$let { ((white ball = x)),((red ball=2x)),((black ball=30−3x)) :} so the number of way to take 2 white balls and 1 black ball is C_2 ^( x) .C_1 ^( 30−3x) = 108 ⇔ ((x^2 −x)/2).(30−3x) = 108 (x^2 −x)(x−10) + 72 = 0 ⇒ x^3 −11x^2 +10x+72 = 0 ⇒(x−4)(x−9)(x+2) = 0 case(1) for x = 4 → { ((red ball = 8)),((black ball=18)) :} so △(b,r) = 18−8 = 10 case(2) for x=9 → { ((red ball=18)),((black ball=3)) :} so △(b,r) = 15.$
	$l e t {\begin{cases} w h i t e b a l l = x \\ r e d b a l l = 2 x \\ b l a c k b a l l = 30 - 3 x \end{cases}$ $s o t h e n u m b e r o f w a y t o t a k e$ $2 w h i t e b a l l s a n d 1 b l a c k b a l l i s$ $C_{2}^{x} . C_{1}^{30 - 3 x} = 108$ $\Leftrightarrow \frac{x^{2} - x}{2} . (30 - 3 x) = 108$ $(x^{2} - x) (x - 10) + 72 = 0$ $\Rightarrow x^{3} - 11 x^{2} + 10 x + 72 = 0$ $\Rightarrow (x - 4) (x - 9) (x + 2) = 0$ $c a s e (1) f o r x = 4 \to {\begin{cases} r e d b a l l = 8 \\ b l a c k b a l l = 18 \end{cases}$ $s o △ (b, r) = 18 - 8 = 10$ $c a s e (2) f o r x = 9 \to {\begin{cases} r e d b a l l = 18 \\ b l a c k b a l l = 3 \end{cases}$ $s o △ (b, r) = 15 .$
	Commented by bemath last updated on 04/Sep/20

	$t h a n k s s i r, a f a r m e r m a t h$
	Commented by john santu last updated on 04/Sep/20
	��
	Commented by Rasheed.Sindhi last updated on 04/Sep/20

	$M a t h i s a f a r m .$ $A l g e b r a, G e o m e t r y, C a l c u l u s,$ $T r i g o n o m e t r y, N u m b e r T h e o r y$ $e t c a r e v a r i o u s k i n d s o f c r o p s$ $AND$ $The farmer is Mr John Santu .$ $Bemath, I & others are$ $consumers!$ $Work hard Mr Santu, b e c a u s e$ $c o n s u m p t i o n i s r a i s i n g h i g h! :)$
	Commented by john santu last updated on 04/Sep/20

	$h a h a h a . . . t h a n k y o u s i r .$
	Commented by bemath last updated on 04/Sep/20

	$a g r e e s i r$
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