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In how many ways can three distinct numbers be selected from the set {1; 2; 3;...; 9} if the product of these numbers is divisible by 21?

Sagot :

AnneCidn
A number may be divisible by 21 if and only if it is divisible by 3 and 7. If we remove these numbers from the set, we will have 7 numbers left. All of these may be used as the third number to satisfy the condition (also, we can not re-use 3 and 7 because we need distinct numbers for each selection). Therefore, there are 7 ways to select three numbers whose product is divisible by 21.


We may also think of it this way. Three lines represent the three numbers to be selected. Since, we need 3 and 7, there will only be one line left. This will be allotted to the remaining 7 numbers from the set.
 3    7    ___

{3,7,1} , {3,7,2} , {3,7,4} and so on. This will result to a total of 7 subsets.
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