Suppose that 7 blue balls , 8 red balls and 9 green balls should be put into three boxes labeled 1,2 and 3, so that any box contains at least one balls of each colour. How many ways can this arrangement be done?


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Question Number 119747 by benjo_mathlover last updated on 26/Oct/20

$S u p p o s e t h a t 7 b l u e b a l l s, 8 r e d b a l l s a n d 9 g r e e n$ $b a l l s s h o u l d b e p u t i n t o t h r e e b o x e s l a b e l e d$ $1, 2 a n d 3, s o t h a t a n y b o x c o n t a i n s a t l e a s t$ $o n e b a l l s o f e a c h c o l o u r . H o w m a n y w a y s$ $c a n t h i s a r r a n g e m e n t b e d o n e ?$
	Answered by mr W last updated on 26/Oct/20

	$t o d i v i d e 7 b l u e b a l l s i n t o 3 g r o u p s,$ $u s i n g s t a r s & b a r s m e t h o d,$ $◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻ ∣ ◻$ $t h e r e a r e C_{2}^{6} w a y s .$ $s i m i l a r l y t o d i v i d e 8 r e d b a l l s i n t o$ $3 g r o u p s t h e r e a r e C_{2}^{7} w a y s, a n d t o$ $d i v i d e 9 g r e e n b a l l s i n t o 3 g r o u p s$ $t h e r e a r e C_{2}^{8} w a y s, s o t o t a l l y w e h a v e$ $C_{2}^{6} \times C_{2}^{7} \times C_{2}^{8} = 8820 w a y s .$
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