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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-arrangeme




Question Number 119747 by benjo_mathlover last updated on 26/Oct/20
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?
$${Suppose}\:{that}\:\mathrm{7}\:{blue}\:{balls}\:,\:\mathrm{8}\:{red}\:{balls}\:{and}\:\mathrm{9}\:{green} \\ $$$${balls}\:{should}\:{be}\:{put}\:{into}\:{three}\:{boxes}\:{labeled} \\ $$$$\mathrm{1},\mathrm{2}\:{and}\:\mathrm{3},\:{so}\:{that}\:{any}\:{box}\:{contains}\:{at}\:{least} \\ $$$${one}\:{balls}\:{of}\:{each}\:{colour}.\:{How}\:{many}\:{ways} \\ $$$${can}\:{this}\:{arrangement}\:{be}\:{done}? \\ $$
Answered by mr W last updated on 26/Oct/20
to divide 7 blue balls into 3 groups,  using stars & bars method,  □∣□∣□∣□∣□∣□∣□  there are C_2 ^6  ways.  similarly to divide 8 red balls into  3 groups there are C_2 ^7  ways, and to  divide 9 green balls into 3 groups  there are C_2 ^8  ways, so totally we have  C_2 ^6 ×C_2 ^7 ×C_2 ^8 =8820 ways.
$${to}\:{divide}\:\mathrm{7}\:{blue}\:{balls}\:{into}\:\mathrm{3}\:{groups}, \\ $$$${using}\:{stars}\:\&\:{bars}\:{method}, \\ $$$$\square\mid\square\mid\square\mid\square\mid\square\mid\square\mid\square \\ $$$${there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{6}} \:{ways}. \\ $$$${similarly}\:{to}\:{divide}\:\mathrm{8}\:{red}\:{balls}\:{into} \\ $$$$\mathrm{3}\:{groups}\:{there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{7}} \:{ways},\:{and}\:{to} \\ $$$${divide}\:\mathrm{9}\:{green}\:{balls}\:{into}\:\mathrm{3}\:{groups} \\ $$$${there}\:{are}\:{C}_{\mathrm{2}} ^{\mathrm{8}} \:{ways},\:{so}\:{totally}\:{we}\:{have} \\ $$$${C}_{\mathrm{2}} ^{\mathrm{6}} ×{C}_{\mathrm{2}} ^{\mathrm{7}} ×{C}_{\mathrm{2}} ^{\mathrm{8}} =\mathrm{8820}\:{ways}. \\ $$

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