Answer the following questions independently of each other
In how many ways can 12 identical bouquets be kept in 3 distinct crates so that the smallest box has at least 2 bouquets, the largest box has at least 4 bouquets and the third one has at least 3 bouquets ?

Options

A. 10
B. 20
C. 21
D. 32

Answer

To solve the problem of distributing 12 identical bouquets into 3 distinct crates with the given constraints, we can follow these steps:

### Step 1: Understand the constraints
We have 3 crates:
- Crate 1 (smallest box) must have at least 2 bouquets.
- Crate 2 (largest box) must have at least 4 bouquets.
- Crate 3 must have at least 3 bouquets.

### Step 2: Allocate the minimum required bouquets
First, we allocate the minimum number of bouquets to each crate based on the constraints:
- Crate 1: 2 bouquets
- Crate 2: 4 bouquets
- Crate 3: 3 bouquets

Now, let's calculate the total number of bouquets allocated:
\[ 2 + 4 + 3 = 9 \]

### Step 3: Calculate remaining bouquets
Next, we find out how many bouquets are left after the minimum allocation:
\[ 12 - 9 = 3 \]
So, we have 3 bouquets remaining to distribute freely among the 3 crates.

### Step 4: Use the stars and bars method
To distribute the remaining 3 bouquets into 3 crates, we can use 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, \( n = 3 \) (remaining bouquets) and \( r = 3 \) (crates).

### Step 5: Substitute values into the formula
Now we substitute the values into the formula:
\[ \binom{3 + 3 - 1}{3 - 1} = \binom{5}{2} \]

### Step 6: Calculate the combination
Now we calculate \( \binom{5}{2} \):
\[ \binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10 \]

### Conclusion
Thus, the total number of ways to distribute the 12 identical bouquets into the 3 distinct crates, given the constraints, is **10**.