Question-133642

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) \dots

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} + \dots)}^{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} + \dots)}^{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}

Facebook Tweet Pin