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

Options

A. A. 250
B. B. 125
C. C. 150
D. D. 300

Answer

To find the number of different ways to display three varieties of perfumes in five places in a showcase, we can follow these steps:

### Step 1: Understand the Problem
We have three varieties of perfumes (let's call them A, B, and C) and five places in a row to display them. Since there is a large number of bottles of each variety, we can use any variety in any place.

### Step 2: Calculate the Total Combinations
Since there are no restrictions on how we can fill the places, each of the 5 places can be filled with any of the 3 varieties. Therefore, for each place, we have 3 choices.

The total number of ways to fill the 5 places is given by:
\[ \text{Total Ways} = 3^5 \]

### Step 3: Calculate \(3^5\)
Now we calculate \(3^5\):
\[ 3^5 = 3 \times 3 \times 3 \times 3 \times 3 = 243 \]

### Step 4: Consider Restrictions
Next, we consider the restrictions mentioned in the video transcript. We need to subtract the cases where all places are filled with only one variety and where only two varieties are used.

#### Case 1: All Places with One Variety
If all 5 places are filled with the same variety, we can choose any of the 3 varieties. Thus, there are:
\[ 3 \text{ ways (A, B, or C)} \]

#### Case 2: All Places with Two Different Varieties
To find the number of ways to fill the places with exactly two different varieties, we first choose 2 varieties from the 3 available. The number of ways to choose 2 varieties from 3 is given by:
\[ \binom{3}{2} = 3 \]
For each pair of varieties chosen, we can fill the 5 places in such a way that at least one of each variety is used. The total arrangements for 5 places using 2 varieties can be calculated as:
\[ 2^5 - 2 = 32 - 2 = 30 \]
(The subtraction of 2 accounts for the cases where all places are filled with only one of the two chosen varieties.)

### Step 5: Combine All Cases
Now we combine all the cases:
1. Total arrangements without restriction: \(3^5 = 243\)
2. Subtract arrangements with one variety: \(3\)
3. Subtract arrangements with exactly two varieties: \(3 \times 30 = 90\)

So, the final calculation is:
\[ \text{Total Ways} = 243 - 3 - 90 = 150 \]

### Final Answer
The number of different ways of displaying the three varieties of perfumes in the showcase is:
\[ \boxed{150} \]