<s>[INST] What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$ , and $a \leq b \leq c $? 
What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$, and $a \leq b \leq c$?

My try: 
The prime factorization of $1000$ is $2^3\cdot 5^3$ 
$a\cdot b \cdot c = 2^3\cdot 5^3$
$a=2^{a_1}\cdot 5^{a_2}$
$b=2^{b_1}\cdot 5^{b_2}$
$c=2^{c_1}\cdot 5^{c_2}$
$abc=2^{a_1+b_1+c_1}\cdot 5^{a_2+b_2+c_2}=2^3\cdot 5^3 $
$a_1+b_1+c_1=3$
How many ways are there such that $a_1+b_1+c_1=3$
Star's and Bar's  method: -
Number of ways to chose $2$ separators($0s$) in a string of $5 $$ = {5\choose 2 }=10$
$N(a_1+b_1+c_1=3)=10$
Similarly,  $N(a_2+b_2+c_2=3)=10$
$N(abc=1000)=10\cdot 10=\boxed{100}$
Is that okay ? Please write down any notes 
 [/INST] For all $a,b,c$ you got 100 but that didn't take into account $a \le b \le c$.
So count $a=b=c$ that is $a=b=c=10$ you counted that once.  The remaining $99$ were overcounted.
Consider $a=b, c\ne a$.  Then $a=b= 2^j5^k; c=2^{3-2j}3^{3-2j}$ so there are $3$ such cases ($j = 0,1; k = 0,1$ but not $j=k=3-2j=3-2k= 1$).  You counted each of those $3$.  So that accounts for $9$ when you should have only counted $3$.  You have $90$ more to account for. 
These are $a,b,c$ distinct.  You counted each of these $6$ times when you should have counted them once.  So you counted $90$ when you should have counted $15$.
So the number should be $15 + 3 + 1 = 19$.
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