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 
 A: 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 $15 + 3 + 1 = 19$.
A: Your computation of $N=10$ is correct and $100$ is the number of ordered triples that have product $1000$.  You have failed to account for the condition that $a \le b \le c$.  All of the unordered triples that have three distinct elements have shown up six times, so you have overcounted.  Those that have two or three equal elements have been counted differently.  Keep going.
