Filling rows with 3 variety of objects 
A shopkeeper sells three varieties of perfumes and he has a large number of bottles of the same size of each variety in his stock. There are $5$ places in a row in his showcase.The number of different ways of displaying the three varieties of perfumes in the show case is

Method 1
This is solved in my reference as, check Q.18
$$
{}^{3}C_{1}.{}^{2}C_{2}.\frac{5!}{3!}+{}^{3}C_{2}.{}^{1}C_{1}\frac{5!}{2!2!}=60+90=150
$$
Method 2
And also as follows, by considering $5$ identical objects are grouped into $3$ distinct bins, where empty groups are allowed, check link
$$
{}^{n+r-1}C_{r-1}={}^{5+3-1}C_{3-1}={}^{7}C_{2}=21
$$
Method 3
And why is it not
$$
3^5=243
$$
Method 4
If the order matters and atleast one variety is present in all times, why is it not
$$5!.{}^{5-1}C_{3-1}.5!={}^{4}C_{2}.5!=6.5!=6!$$
where ${}^{5-1}C_{3-1}$ is number of ways in which $5$ identical objects divided into $3$ groups when empty groups are not allowed.
What is different in each attempts ?
There are other similar problems like this in which I am having kinda same confusion, example

Find the number of ways of selecting $10$ balls out of an unlimited number of identical white, red, and blue balls.
$$
{}^{10+3-1}C_{3-1}={}^{12}C_{2}=\frac{12*11}{2}=66
$$
In a steamer there are stalls for $12$ animals, and there are Cows, Sheep and Calves (not less than $12$ each) ready to be shipped; in how many ways can the shipload be made?
$$
3^{12}
$$

 A: In the first problem, we are arranging the objects.  In the second and third problems, we are selecting the objects, but we are not arranging them.  The fourth problem is a bit ambiguous, but I suspect the author is interested in which animals are to be shipped rather than how to arrange them in the stalls.

A shopkeeper sells three varieties of perfumes and he has a large number of bottles of the same size of each variety in his stock. There are $5$ places in a row in his showcase.  The number of different ways of displaying the three varieties of perfumes in the show case is

In this problem, the order in which the perfume bottles are arranged matters.  Also, the author is assuming all three types of perfume are displayed.  There are two ways to express $5$ as the sum of three positive integers:
\begin{align*}
5 & = 3 + 1 + 1\\
  & = 2 + 2 + 1
\end{align*}
Three bottles of one perfume and one bottle apiece of each of the others:  There are three choices for the type of perfume that will appear three times, $\binom{5}{3}$ ways to choose three positions for those bottles, and two ways to decide which of the other perfumes will occupy the leftmost open position. The other perfume must occupy the final open position. Hence, there are 
$$\binom{3}{1}\binom{5}{3}\binom{2}{1}$$
such displays.
Two bottles apiece of two types of perfume and one bottle of the other: There are three choices the type of perfume that will appear once, five ways to choose its position, and $\binom{4}{2}$ ways to choose two of the remaining four positions for bottles of the remaining type of perfume that appears first among those two types on the inventory list.  The final two positions must be filled by the other type of perfume.  Hence, there are 
$$\binom{3}{1}\binom{5}{1}\binom{4}{2}$$
such displays.
Since the two cases are mutually exclusive and exhaustive, the number of such displays is 
$$\binom{3}{1}\binom{5}{3}\binom{2}{1} + \binom{3}{1}\binom{5}{1}\binom{4}{2}$$

In how many ways can five identical objects be grouped in three distinct bins, where empty bins are allowed?

In this case, what matters is how many objects are placed in each bin.  Let $x_1$, $x_2$, $x_3$ denote the number of objects placed in the first, second, and third bins, respectively.  Since a total of five objects are placed in the three bins,
$$x_1 + x_2 + x_3 = 5$$
is an equation in the nonnegative integers.  A particular solution corresponds to the placement of $3 - 1 = 2$ addition signs in a row of five ones.  For instance,
$$1 1 1 + 1 + 1$$
corresponds to the solution $x_1 = 3$, $x_2 = 1$, $x_3 = 1$, while
$$+ + 1 1 1 1 1$$
corresponds to the solution $x_1 = 0$, $x_2 = 0$, $x_3 = 5$.  The number of such distributions is equal to the number of solutions of the equation $x_1 + x_2 + x_3 = 5$ in the nonnegative integers, which is 
$$\binom{5 + 3 - 1}{3 - 1} = \binom{7}{2}$$
since we must select which two of the seven positions required for five ones and two addition signs will be filled with addition signs.
