Question

A shopkeeper sells three varieties of fruit juice. He has a large number of bottles of same size of each variety. The number of different ways of displaying all the three varieties on the shelf with 5 places in a row and each display must have at least one bottle of each variety is

• A. 150
• B. 120
• C. 60
• D. 90

Hint:

When identical objects of different varieties are arranged with at least one of each type, first distribute the counts and then arrange using multinomial formula.

Answer 1

The Correct Option is A

Step 1: Understand the arrangement condition.
There are 5 places in a row and 3 varieties of fruit juice bottles.
Each display must contain at least one bottle of each variety.

Step 2: Represent the number of bottles of each variety.
Let the three varieties be \(A\), \(B\), and \(C\).
Let their numbers in a display be:
\[ x+y+z=5 \] where
\[ x\geq 1,\quad y\geq 1,\quad z\geq 1. \]

Step 3: Find possible distributions.
Since 5 bottles are to be divided among 3 varieties with each variety appearing at least once, the possible distributions are:
\[ (3,1,1) \] and
\[ (2,2,1). \]

Step 4: Count arrangements for distribution \( (3,1,1) \).
First choose which variety appears 3 times.
This can be done in:
\[ 3 \] ways.
For each such choice, arrangements of 5 bottles are:
\[ \frac{5!}{3!1!1!}=20. \]
So total arrangements for this case are:
\[ 3 \times 20 = 60. \]

Step 5: Count arrangements for distribution \( (2,2,1) \).
First choose which variety appears only once.
This can be done in:
\[ 3 \] ways.
For each such choice, arrangements of 5 bottles are:
\[ \frac{5!}{2!2!1!}=30. \]
So total arrangements for this case are:
\[ 3 \times 30 = 90. \]

Step 6: Add both cases.
\[ \text{Total ways}=60+90. \]
\[ =150. \]

Step 7: Final conclusion.
Thus, the number of different displays is 150.
Final Answer:
\[ \boxed{150} \]