# 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?

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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.

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.

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