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

Options

A. 6
B. 50
C. 150
D. 90

Answer

To solve the problem of displaying three varieties of perfumes in five places, we will consider two distinct cases based on how many times each variety is displayed.

### Step-by-Step Solution:

**Step 1: Understand the Problem**
We have three varieties of perfumes (let's call them A, B, and C) and five places to fill in a showcase. We need to find the total number of different ways to display these perfumes.

**Step 2: Case 1 - One Variety Appears Three Times**
In this case, we will choose one variety to display three times, and the other two varieties will each occupy one of the remaining two places.

1. **Choose the Variety**: We can choose any one of the three varieties (A, B, or C) to be displayed three times. This can be done in \(3\) ways.
2. **Choose the Remaining Varieties**: The remaining two varieties will occupy the last two places. There is only \(1\) way to choose the remaining two varieties since we are left with two after selecting one.
3. **Arrange the Bottles**: The arrangement of the bottles in the five places can be calculated using the formula for permutations of multiset:
\[ \frac{5!}{3!} = \frac{120}{6} = 20 \]
Here, \(5!\) is the total arrangements of 5 bottles, and \(3!\) accounts for the three identical bottles of the chosen variety.

4. **Total for Case 1**:
\[ \text{Total for Case 1} = 3 \times 1 \times 20 = 60 \]

**Step 3: Case 2 - Two Varieties Appear Twice**
In this case, we will choose two varieties to display twice each, and the third variety will occupy the last place.

1. **Choose the Two Varieties**: We can choose any two varieties from the three available. This can be done in \( \binom{3}{2} = 3 \) ways.
2. **Choose the Remaining Variety**: The remaining variety will occupy the last place, which can only be done in \(1\) way.
3. **Arrange the Bottles**: The arrangement of the bottles in the five places can be calculated as:
\[ \frac{5!}{2! \times 2!} = \frac{120}{4} = 30 \]
Here, \(2!\) accounts for the two identical bottles of the first chosen variety and another \(2!\) for the second chosen variety.

4. **Total for Case 2**:
\[ \text{Total for Case 2} = 3 \times 1 \times 30 = 90 \]

**Step 4: Combine Both Cases**
Now, we add the totals from both cases to find the overall number of ways to display the perfumes:
\[ \text{Total Ways} = \text{Total for Case 1} + \text{Total for Case 2} = 60 + 90 = 150 \]

### Final Answer:
The total number of different ways of displaying the three varieties of perfumes in the showcase is **150**.