Number of positive integral solution of product $x_{1} \cdot x_{2} \cdot x_{3}\cdot x_{4}\cdot x_{5}=1050$ is The number of positive integral solution of product $x_{1} \cdot x_{2} \cdot x_{3}\cdot x_{4}\cdot  x_{5}=1050$ is

$\bf{My\; Try}::$ Given $x_{1}\cdot x_{2}\cdot x_{3}\cdot x_{4}\cdot x_{5} = 2 \times 3 \times 5^2 \times 7$.

Now Let we have $5$ Different Boxes $x_{1}\;,x_{2}\;,x_{3}\;,x_{4}\;,x_{5}$ and $5$ balls in 

which prints a numbers $2\;,3\;,5\;,5\;,7$.

Now We have to put balls into boxes , So we will form different cases.

$\bullet \;$ If all the balls are in same boxes(boxes contain all balls or no balls) 

Then no. of ways $ = 5\times 5 \times 5\times 5 \times 5 = 5^5$

$\bullet \; $ If all balles are in different boxes, Then no. of ways $\displaystyle  = $

Now I did not understand how can i find in $\bf{(II)}$ cases, 

Help me Thanks
You can put $2,3,7$ in the boxes in five ways each. The fives can be placed in $5+\binom 52$ ways.

$5$ represents both of the $5$s in the same box. $\binom 52$ represents them in different boxes.

So you should get $5\times 5 \times 5 \times 15$ ways.



It appears that the first case in the question, where the $5$s are in the same box should give the result $5^4$ rather than $5^5$ because the pair of $5$s acts as a single unit rather than two separate units.

If the $5$s are in separate boxes, they are indistinguishable, and can be swapped without changing the factorisation. The first $5$ can be placed in any of $5$ boxes, and the second in any of the remaining $4$ boxes, but this counts every possibility twice. So the number of ways of placing the $5$s is $\cfrac {5\times 4}2=10$. This second case counts for $5^3\times 10$ possibilities.

The number of ways of choosing $r$ places out of a possible $n$ is so commonly encountered that it is given the (modern) symbol $\binom nr=\frac {n!}{r!(n-r)!}$. Apologies if this confused you, but it is useful to know, as you are likely to encounter it frequently, especially if you ask similar questions on this site.
Each box can contain $5$ elements so it will be toatal $5^5$ cases I think.