in-how-many-ways-we-can-distribute-6-distinct-balls-in-3-identical-boxes-

Question Number 212522 by Nadirhashim last updated on 16/Oct/24

i n h o w m a n y w a y s w e

c a n d i s t r i b u t e 6 d i s t i n c t

b a l l s i n 3 i d e n t i c a l b o x e s

Answered by mehdee7396 last updated on 16/Oct/24

6 / 0 / 0 o r 5 / 1 / 0 o r 4 / 2 / 0 o r 4 / 1 / 1 o r 3 / 3 / 0 / o r 3 / 2 / 1 o r 2 / 2 / 2

C (6, 6) + C (5, 1) C (1, 1) + C (6, 4) C (2, 2) + C (6, 4) C (2, 1) C (1, 1) + C (6, 3) C (3, 3) + C (6, 3) C (3, 2) C (1, 1) + C (6, 2) C (4, 2) C (2, 2)

= 275 ✓

Commented by mr W last updated on 21/Oct/24

{i t}^{'} s w r o n g s i r .

t h e n u m b e r o f w a y s t o d i s t r i b u t e n

d i s t i n c t o b j e c t s i n t o

n_{1} g r o u p s w i t h m_{1} o b j e c t s i n e a c h a n d

n_{2} g r o u p s w i t h m_{2} o b j e c t s i n e a c h

e t c .

(n = n_{1} m_{1} + n_{2} m_{2} + \dots + n_{r} m_{r})

i s \frac{n!}{\underset{k = 1}{\prod^{r}} {(m_{k}!)}^{n_{k}} n_{k}!}

t h e r e f o r e f o r e x a m p l e

3 / 2 / 1 \Rightarrow \frac{6!}{3! 2! 1!} = 60 = C (6, 3) C (3, 2) C (1, 1)

2 / 2 / 2 \Rightarrow \frac{6!}{{(2!)}^{3} 3!} = 15 \neq C (6, 2) C (4, 2) C (2, 2)

Answered by mr W last updated on 16/Oct/24

i f t h e b o x e s m a y n o t b e e m p t y :

S (6, 3) = 90 ✓

i f t h e b o x e s m a y b e e m p t y :

S (6, 3) + S (6, 2) + S (6, 1) = 90 + 31 + 1 = 122 ✓

Answered by mr W last updated on 17/Oct/24

6 / 0 / 0 \Rightarrow 1 w a y

5 / 1 / 0 \Rightarrow C_{1}^{6} = 6 w a y s

4 / 2 / 0 \Rightarrow C_{2}^{6} = 15 w a y s

3 / 3 / 0 \Rightarrow C_{3}^{6} / 2 = 10 w a y s

4 / 1 / 1 \Rightarrow \frac{6!}{4! {(1!)}^{2} 2!} = 15 w a y s

3 / 2 / 1 \Rightarrow \frac{6!}{3! 2! 1!} = 60 w a y s

2 / 2 / 2 \Rightarrow \frac{6!}{{(2!)}^{3} 3!} = 15 w a y s

t o t a l l y : 122 w a y s ✓

Facebook Tweet Pin