Number-of-distributions-of-n-different-articles-to-r-different-boxes-so-as-1-empty-box-allowed-2-empty-box-not-allowed-with-proof-kindly-help-me-

Question Number 195538 by SLVR last updated on 04/Aug/23

N u m b e r o f d i s t r i b u t i o n s o f

n d i f f e r e n t a r t i c l e s t o r d i f f e r e n t b o x e s

s o a s 1) e m p t y b o x a l l o w e d

2) e m p t y b o x n o t a l l o w e d

w i t h p r o o f \dots k i n d l y h e l p m e

Commented by SLVR last updated on 04/Aug/23

s i r \dots k i n d l y g i v e p r o o f

Commented by SLVR last updated on 05/Aug/23

s i r M r . W . . o r \dots a n y o n e c a n

h e l p m e \dots i n e e d b i t u r g e n t l y

Commented by mr W last updated on 05/Aug/23

u s i n g g e n e r a t i n g f u n c t i o n m e t h o d

w e c a n s e e t h a t t h e r e s u l t i s t h e

c o e f f i c i e n t o f t e r m x^{n} i n e x p a n s i o n

1) n! {(1 + \frac{x}{1} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots)}^{r}, i . e . n! e^{r x}

2) n! {(\frac{x}{1} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \dots)}^{r}, i . e . n! {(e^{x} - 1)}^{r}

Commented by SLVR last updated on 05/Aug/23

T h a n k s P r o f . W \dots s o k i n d o f y o u

Commented by mr W last updated on 05/Aug/23

a l i t t l e e x p l a n a t i o n :

s a y b o x 1 h a s k_{1} o b j e c t s, b o x 2 h a s k_{2}

o b j e c t s, e t c . w i t h k_{1} + k_{2} + \dots k_{r} = n a n d

k_{1}, k_{2}, \dots, k_{r} a r e f i x e d v a l u e s .

t o d i s t r i b u t e n o b j e c t s i n r b o x e s i n

t h i s w a y t h e r e a r e \frac{n!}{(k_{1})! \times (k_{2})! \times \dots (k_{r})!}

p o s s i b i l i t i e s .

s i n c e k_{1}, k_{2}, \dots, k_{r} c a n b e v a r i a b l e v a l u e s,

w e c a n c o n s t r u c t a g e n e r a t i n g

f u n c t i o n l i k e

n! {(\frac{x}{1!} + \frac{x^{2}}{2!} + \dots + \dots)}^{r} i f e a c h b o x m u s t

g e t a t l e a s t a n o b j e c t, o r

n! {(1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \dots + \dots)}^{r} i f e a c h b o x m a y

a l s o b e e m p t y .

f o r 1) t h e f o r m u l a i s r^{n} .

f o r 2) t h e f o r m u l a i s

r^{n} - C_{1}^{r} {(r - 1)}^{n} + C_{2}^{r} {(r - 2)}^{n} - \dots + {(- 1)}^{n - 1} C_{r - 1}^{r}

Commented by mr W last updated on 05/Aug/23

e x a m p l e :

w e h a v e 10 f o r e i g n s t u d e n t s w h o

s h o u l d b e d i s t r i b u t e d t o 3 s c h o o l s .

i n h o w m a n y w a y s c a n t h i s b e d o n e ?

1) 3^{10} = 59049 w a y s

2) 3^{10} - 3 \times 2^{10} + 3 \times 1^{10} = 55980 w a y s

Commented by mr W last updated on 05/Aug/23

Answered by MM42 last updated on 05/Aug/23

i f n ⩾ r

l e t x_{i} = n u m b e r a r t i c l e s i n t h e i^{t h} b o x

1) \Rightarrow x_{1} + x_{2} + \dots + x_{r} = n; x_{i} ⩾ 0

a n s = (\begin{matrix} n + r - 1 \\ n \end{matrix}) = \frac{(n + r - 1)!}{n! (r - 1)!}

2) \Rightarrow x_{1} + x_{2} + \dots + x_{r} = n; x_{i} ⩾ 1

\Rightarrow y_{1} + y_{2} + \dots + y_{r} = n - r; y_{i} ⩾ 0

a n s = (\begin{matrix} n - 1 \\ n - r \end{matrix}) = \frac{(n - 1)!}{(n - r)! (r - 1)!}

i f n < r t h e n n o a l l b o x c a n b e f u l l .

Commented by mr W last updated on 05/Aug/23

w h a t y o u s o l v e d i s f o r t h e c a s e t h a t

t h e o b j e c t s a r e i d e n t i c a l . b u t t h e

q u e s t i o n s a i d t h a t b o t h t h e b o x e s

a n d t h e o b j e c t s a r e d i f f e r e n t .

Commented by MM42 last updated on 08/Aug/23

y e s . y o u a r e r i g h t . i d i d n o t n o t i c e

t h e d i f f r e n c e b e t w e e n t h e s u b j e c t s .

t h a n k s f o r y o u

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