Number of distributions of 12 different things be taken to 3different boxes so as 1)5 things in 1st box exactly 2)5 things in any one box?


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Question Number 179738 by SLVR last updated on 01/Nov/22

$N u m b e r o f d i s t r i b u t i o n s o f$ $12 d i f f e r e n t t h i n g s b e t a k e n t o$ $3 d i f f e r e n t b o x e s s o a s$ $1) 5 t h i n g s i n 1 s t b o x e x a c t l y$ $2) 5 t h i n g s i n a n y o n e b o x ?$
	Answered by SLVR last updated on 01/Nov/22

	$I a m g e t t i n g 12_{C_{5}} \times 2^{7} f o r$ $1 s t a n d b u t 2 n d$ $q u e s t i o n . =>> p l e a s e$
	Answered by mr W last updated on 03/Nov/22

	$1)$ $5 t h i n g s f o r t h e 1 s t b o x : C_{5}^{12}$ $7 t h i n g s i n t o t w o b o x e s : 2^{7}$ $\Rightarrow C_{5}^{12} \times 2^{7} = 101376$ $2)$ $n o t c l e a r, w h a t y o u m e a n .$ $a t l e a s t o r a t m o s t o r e x a c t l y 5 t h i n g s$ $i n o n e o f t h e t h r e e b o x e s ?$
	Commented by SLVR last updated on 02/Nov/22

	$s i r . . s o k i n d o f y o u . .$ $h e r e i n 2 n d i n e e d 5 e x a c t l y$ $b u t w h a t i f a) i f a t l e a s t o n e$ $i n 2 n d b o x ?$ $b) i f a t m o s t 7 i n 2 n d b o x$ $p l e a s e s i r P r o f . M r . W$
	Commented by mr W last updated on 03/Nov/22

	$2 a)$ $s a y i n o n e a n d o n l y o n e b o x t h e r e a r e$ $e x a c t l y 5 b a l l s .$ $t o d i v i d e 12 b a l l s i n t o t h r e e b o x e s$ $t h e r e a r e f o l l o w i n g p o s s i b i l i t i e s :$ $5 + 0 + 7 ✓ \Rightarrow \frac{12!}{5! 7!} \times 3!$ $5 + 1 + 6 ✓ \Rightarrow \frac{12!}{5! 6!} \times 3!$ $5 + 2 + 5 \times$ $5 + 3 + 4 ✓ \Rightarrow \frac{12!}{5! 3! 4!} \times 3!$ $\Rightarrow a n s w e r i s$ $\frac{12! 3!}{5!} \times (\frac{1}{7!} + \frac{1}{6!} + \frac{1}{3! 4!}) = 204 336$
	Commented by mr W last updated on 03/Nov/22

	$2 b)$ $s a y i n o n e b o x a t l e a s t 5 b a l l s :$ $12 + 0 + 0 \Rightarrow 1 \times \frac{3!}{2!} = 3$ $11 + 1 + 0 \Rightarrow \frac{12!}{11!} \times 3! = 72$ $10 + 2 + 0 \Rightarrow \frac{12!}{10! 2!} \times 3! = 396$ $10 + 1 + 1 \Rightarrow \frac{12!}{10! 2!} \times 3! = 396$ $9 + 3 + 0 \Rightarrow \frac{12!}{9! 3!} \times 3! = 1 320$ $9 + 2 + 1 \Rightarrow \frac{12!}{9! 2!} \times 3! = 3 960$ $8 + 4 + 0 \Rightarrow \frac{12!}{8! 4!} \times 3! = 2 970$ $8 + 3 + 1 \Rightarrow \frac{12!}{8! 3!} \times 3! = 11 880$ $8 + 2 + 2 \Rightarrow \frac{12!}{8! {(2!)}^{2} 2!} \times 3! = 8 910$ $7 + 5 + 0 \Rightarrow \frac{12!}{7! 5!} \times 3! = 4 752$ $7 + 4 + 1 \Rightarrow \frac{12!}{7! 4!} \times 3! = 23 760$ $7 + 3 + 2 \Rightarrow \frac{12!}{7! 3! 2!} \times 3! = 47 520$ $6 + 6 + 0 \Rightarrow \frac{12!}{{(6!)}^{2} 2!} \times 3! = 2 772$ $6 + 5 + 1 \Rightarrow \frac{12!}{6! 5!} \times 3! = 33 264$ $6 + 4 + 2 \Rightarrow \frac{12!}{6! 4! 2!} \times 3! = 83 160$ $6 + 3 + 3 \Rightarrow \frac{12!}{6! {(3!)}^{2} 2!} \times 3! = 55 440$ $5 + 5 + 2 \Rightarrow \frac{12!}{{(5!)}^{2} 2! 2!} \times 3! = 49896$ $5 + 4 + 3 \Rightarrow \frac{12!}{5! 4! 3!} \times 3! = 166 320$ $t o t a l l y : 496 791$
	Commented by SLVR last updated on 02/Nov/22

	$N o w . . . i g o t t h e c l a r i t y . . s o$ $k i n d o f s i r . . t h a n k s a l o t$
	Commented by mr W last updated on 03/Nov/22

	$g e n e r a t i n g f u n c t i o n s a r e p o w e r f u l$ $t o o l s f o r s o l v i n g s u c h q u e s t i o n s .$ $w e {d o n}^{'} t n e e d t o e n u m e r a t e t h e$ $c o n c r e t e c o m b i n a t i o n s a s i d i d a b o v e .$ $e x a m p l e s :$ $1)$ $c o e f . o f t e r m x^{12} i n$ $12! \times \frac{x^{5}}{5!} \times {(e^{x})}^{2}$ $w h i c h i s 101 376 .$
	Commented by mr W last updated on 03/Nov/22

	Commented by mr W last updated on 03/Nov/22

	$2 a)$ $c o e f . o f t e r m x^{12} i n$ $3 \times 12! \times \frac{x^{5}}{5!} \times {(e^{x} - \frac{x^{5}}{5!})}^{2}$ $w h i c h i s 204 336 .$
	Commented by mr W last updated on 03/Nov/22

	Commented by mr W last updated on 03/Nov/22

	$2 b)$ $w i t h o u t r e s t r i c t i o n : 3^{12} = 531 441 w a y s$ $a l l b o x e s h a v e l e s s t h a n 5 b a l l s, i . e .$ $e a c h h a s 4 b a l l s : \frac{12!}{{(4!)}^{3}} = 34650 w a y s$ $o n e b o x h a s a t l e a s t 5 b a l l s :$ $\Rightarrow 531 441 - 34 650 = 496 791$ $o r u s i n g g e n e r a t i n g f u n c t i o n s :$
	Commented by mr W last updated on 03/Nov/22

	Commented by mr W last updated on 03/Nov/22

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