**How many**

__ordered triples__of pairwise distinct, positive integers**(**

*a***,**

*b***,**

*c***)**

**are there such that**

*abc***=**

**10**

^{6}**?**

**Clarification: Ordered triples -**

**(**

**1**

**,**

**10**

**,**

**10**

^{5}**)**

**and**

**(**

**10**

^{5}**,**

**10**

**,**

**1**

**)**

**count as different solutions.**

**Clarification: Pairwise distinct - a set of integers are pairwise distinct if no two of them are the same. For example, the set**

**(**

**1**

**,**

**1**

**,**

**10**

^{6}**)**

**is NOT pairwise distinct, hence does not count as a solution.**

Set

*a*=2^{a}^{1}⋅5^{a}^{2},*b*=2^{b}^{1}⋅5^{b}^{2},*c*=2^{c}^{1}⋅5^{c}^{2}. We first count the number of triples without restriction. We seek the number of non-negative integer solutions to*a*_{1}+*b*_{1}+*c*_{1}=6,*a*_{2}+*b*_{2}+*c*_{2}=6. This is equivalent to arranging 6 1's with 2 0's (dividing blocks), thus there are (6+22) ways to solve either of the two equations. By the rule of product, there are (82)⋅(82)=784 solutions in total.
If

*a*=*b*, we seek the number of non-negative integer solutions to 2*a*_{1}+*c*_{1}=6,2*a*_{2}+*c*_{2}=6. For each equation, there are 4 solutions, corresponding to*a**i*=0,1,2,3. By the product rule, there are 4⋅4=16 solutions in all. The cases*b*=*c*and*c*=*a*are exactly the same. If*a*=*b*=*c*, there is exactly 1 solution.
Let

*A*,*B*, and*C*be the set of triples where*b*=*c*,*c*=*a*, and*a*=*b*, respectively. By the Principle of Inclusion and Exclusion, |*A*∪*B*∪*C*|=|*A*|+|*B*|+|*C*|−|*A*∩*B*|−|*B*∩*C*|−|*C*∩*A*|+|*A*∩*B*∩*C*|=16+16+16−1−1−1+1=46. Hence, the number of triples with pairwise distinct values is 784−46=738.
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