Answer the following questions independently of each other
In how many ways can 12 identical bouquets be kept in 3 distinct crates so that at least 2 bouquets must be there in each bouquet ?

Options

A. 28
B. 24
C. 42
D. 36

Answer

To solve the problem of distributing 12 identical bouquets into 3 distinct crates with the condition that at least 2 bouquets must be in each crate, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have 12 identical bouquets and we need to distribute them into 3 distinct crates. Each crate must contain at least 2 bouquets.

2. **Initial Distribution**: Since each crate must have at least 2 bouquets, we start by placing 2 bouquets in each crate. This means we will initially use:
\[ 2 \text{ bouquets} \times 3 \text{ crates} = 6 \text{ bouquets} \]
After this initial distribution, we have:
\[ 12 \text{ total bouquets} - 6 \text{ bouquets used} = 6 \text{ bouquets remaining} \]

3. **Distributing Remaining Bouquets**: Now, we need to distribute the remaining 6 bouquets among the 3 crates. This is a classic problem of distributing indistinguishable objects (bouquets) into distinct boxes (crates).

4. **Using the Stars and Bars Theorem**: The problem can be solved using the "stars and bars" theorem. The formula for distributing \( n \) identical items into \( r \) distinct groups is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 6 \) (remaining bouquets) and \( r = 3 \) (crates).

5. **Applying the Formula**:
\[ \text{Number of ways} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} \]

6. **Calculating the Binomial Coefficient**:
\[ \binom{8}{2} = \frac{8!}{2!(8-2)!} = \frac{8 \times 7}{2 \times 1} = \frac{56}{2} = 28 \]

7. **Final Answer**: Therefore, the total number of ways to distribute the 12 identical bouquets into 3 distinct crates, ensuring at least 2 bouquets in each crate, is **28**.