The number of positive integral solutions of the equation $x_1x_2x_3x_4x_5=1050$? 
The number of positive integral solutions of the equation
  $x_1x_2x_3x_4x_5=1050$ ?

I prime factorized 1050.Then what to do?
 A: HINT:
$1050 = 2\cdot 3 \cdot 5^2 \cdot 7$. Do for each power of a prime separately. Say, for $y_1y_2 y_3 y_4= 2$ there are $4$ solutions, 
while for $y_1y_2 y_3 y_4= 5^2$ there are $4+ 6 = 10$. From each solution for the separate prime powers one gets a solution for the equation. Thus the number is the product of all these numbers $4 \cdot 4 \cdot 10 \cdot 4 = \ldots $
A: Hint
$1$ is also a positive integer !
Arrange the factors in ascending order:
With four $1's : 1$
With three $1's: (2\cdot x_5), (3\cdot x_5),$ etc  
With two $1's: (2\cdot3\cdot x_5), (2\cdot5\cdot x_5),$ etc
With one $1:(2\cdot3\cdot5\cdot x_5), (2\cdot3\cdot7\cdot x_5),$ etc 
With no $1's: 1$
Work out for each case and add up.
Further hint:
There are $4$ distinct factors. If taken one at a time (or four at a time), there will obviously be $4$,
but if taken $2$ at a time,(or $3$ at a time) there will be $\binom42 +\binom31 = 7$
(the second term is when both the $5's$ occur)
The list is $(2,3) (2,5) (2,7) (3,5) (3,7) (5,5) (5,7)$
Now consider the case where three $1's$ are there.
You can either have the remaining two split $1-4$ or $2-3$
Proceed.

Revised formulation after reading clarification to G.Edgar's query
It now becomes very simple:
The $3$ unique factors can be placed in $5^3$ ways in a row of $5$,
and the $5's$ can be placed in $\binom52$ ways separately, $5$ ways together,
which yields a total of $5^3\cdot15 = 1875$
All places left blank by this process will be filled by a $1$  
A: The answer is 
$$10\cdot5^3+5^4=15\cdot125=1875$$
Think of the $x_i$'s as a row of $5$ boxes into which you're going to throw the prime numbers $2$, $3$, $5$, $5$, and $7$ (whose product is $1050$).  The two $5$'s either go into separate boxes, or they go together.  There are ${5\choose2}=10$ ways they can go into separate boxes and $5$ ways they can go together.  The other primes have $5$ choices each.
A: Let $ x_i=2^{a_i}3^{b_i}5^{c_i}7^{d_i}$ for $1\le i\le5$.
Then $a_i=1$ for exactly one $i$, with the remaining $a_i=0;\;$
and similarly for the $b_i$ and $d_i$.
For the $c_i$ we have two cases:
A) $c_i=2$ for exactly one $i$ and the remaining $c_i=0$, 
$\;\;\;$in which case there are $5^4=625$ possibilities;
B) $c_i=1$ for exactly two values of $i$ and the remaining $c_i=0$;
$\;\;\;$ in which case there are $5^3\binom{5}{2}=1250$ possibilities.
Therefore there are a total of 1875 solutions of this form.
A: The $2$ must divide some $x_i$, so there are $5$ options for which $x_i$ it divides. Similarly for the $3$ and the $7$. 
The $5$s can be distributed amongst the $x_i$ in $5+\binom{5}{2}=15$ ways, counting where $25$ divides some $x_i$ and where the $5$s are separated as dividing two distinct $x_i$. 
So there are $5\cdot5\cdot5\cdot15=1875$ solutions.
