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Question Number 133642 by liberty last updated on 23/Feb/21

Commented by liberty last updated on 23/Feb/21

$dear Mr W, Mr Edwin and all master$
	Answered by bobhans last updated on 23/Feb/21

	${I t}^{'} s t h e s a m e a s t h e n u m b e r o f w a y s$ $o f p l a c i n g 2 i d e n t i c a l d i v i d e r s i n$ $a l i n e o f 12 i d e n t i c a l 12 b a l l s, w h i c h$ $i s t h e s a m e a s t h e n u m b e r o f w a y s$ $o f c h o o s i n g 2 i t e m s f r o m 14$ $d i s t i n c t i t e m s = C_{2}^{14} = 91$
	Commented by liberty last updated on 23/Feb/21

	$thank you$
	Answered by mr W last updated on 23/Feb/21

	$◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻$ $i f n o b o x i s e m p t y, t h e n t h e a n s w e r i s$ $C_{3 - 1}^{12 - 1} = C_{2}^{11} = 55$
	Commented by liberty last updated on 23/Feb/21

	$how you got C_{2}^{11} sir ?$ $let the boxes namely A, B and C$ $(1, 1, 10), (1, 2, 9), (1, 3, 8), (1, 4, 7)$ $(1, 5, 6), (2, 2, 8), (2, 3, 7) . . .$
	Commented by mr W last updated on 23/Feb/21

	$a s s h o w n a b o v e, w e h a v e 12 b a l l s ◻, t o$ $d i v i d e t h e m i n t o 3 g r o u p s w e o n l y$ $n e e d t o p l a c e 2 b a r s ∣ a m o n g t h e m .$ $t h e r e a r e 11 p o s s i b l e p o s i t i o n s ∣ t o$ $p l a c e t h e s e 2 b a r s, s o t h e a n s w e r i s$ $C_{2}^{11} . t h i s m e t h o d i s c a l l e d ‘ ‘ s t a r s a n d$ ${b a r s}^{″} m e t h o d .$
	Commented by mr W last updated on 23/Feb/21

	$o r a n o t h e r w a y :$ ${(x + x^{2} + x^{3} + . . .)}^{3} = \frac{x^{3}}{{(1 - x)}^{3}} = x^{3} \underset{k = 0}{\sum^{\infty}} C_{2}^{k + 2} x^{k}$ $c o e f . o f x^{12} i s t h e a n s w e r = C_{2}^{9 + 2} = C_{2}^{11}$ $i f t h e b o x e s m a y b e e m p t y, t h e n$ ${(1 + x + x^{2} + x^{3} + . . .)}^{3} = \frac{1}{{(1 - x)}^{3}} = \underset{k = 0}{\sum^{\infty}} C_{2}^{k + 2} x^{k}$ $c o e f . o f x^{12} i s t h e a n s w e r = C_{2}^{12 + 2} = C_{2}^{14} = 91$
	Commented by liberty last updated on 23/Feb/21

	$thank you$
	Commented by mr W last updated on 23/Feb/21

	$b u t t o d i v i d e 12 d i s t i n c t b a l l s i n t o 3$ $d i s t i n c t b o x e s t h e r e a r e 519156 w a y s!$
	Commented by liberty last updated on 23/Feb/21

	$how to find it ?$
	Commented by mr W last updated on 23/Feb/21

	$u s i n g g e n e r a t i n g f u n c t i o n i t i s t h e$ $c o e f f i c i e n t o f x^{12} t e r m i n t h e$ $e x p a n s i o n o f 12! {(e^{x} - 1)}^{3} .$
	Answered by EDWIN88 last updated on 23/Feb/21

	$the question is equivalent to how many$ $positive number integer solution a + b + c = 12$ $= C_{2}^{11} = 55$ $but if the equation how many non negatif number$ $(whole number) are there a + b + c = 100$ $= C_{2}^{14} = 91$
	Commented by liberty last updated on 23/Feb/21

	$thank you$
	Answered by JDamian last updated on 23/Feb/21

	$T h i s c a n b e t h o u g h t a s t h e n u m b e r o f$ $P e r m u t a t i o n s w i t h r e p e t i t i o n o f$ $12 b a l l s + (3 - 1) s e p a r a t o r s$ $∙ ∙ ∙ ◊ ∙ ∙ ∙ ∙ ◊ ∙ ∙ ∙ ∙ ∙$ $\frac{(12 + 3 - 1)!}{12! \cdot (3 - 1)!} = \frac{14!}{12! \cdot 2!} = C_{2}^{14}$
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