Answer the following questions independently of each other
In how many ways can 12 identical bouquets be kept in 3 identical crates ?

Options

A. 20
B. 16
C. 19
D. 25

Answer

To solve the problem of how many ways 12 identical bouquets can be kept in 3 identical crates, we can use the "stars and bars" theorem. Here’s a step-by-step solution:

### Step 1: Understand the problem
We need to distribute 12 identical bouquets into 3 identical crates. Since the crates are identical, we are interested in the different distributions of bouquets rather than the arrangement of crates.

### Step 2: Set up the equation
Let’s denote the number of bouquets in each crate as \( x_1, x_2, \) and \( x_3 \). We need to find the non-negative integer solutions to the equation:
\[ x_1 + x_2 + x_3 = 12 \]

### Step 3: Use the stars and bars method
In the stars and bars method, the "stars" represent the items to be distributed (in this case, the bouquets), and the "bars" represent the dividers between different groups (the crates).

Since we have 12 bouquets (stars) and we want to divide them into 3 groups (crates), we will have 2 bars. The total number of symbols (stars + bars) is:
\[ 12 + 2 = 14 \]

### Step 4: Calculate the combinations
The number of ways to arrange these symbols is given by the combination formula:
\[ \binom{n + r - 1}{r - 1} \]
where \( n \) is the number of items to distribute (12 bouquets) and \( r \) is the number of groups (3 crates).

So, we need to calculate:
\[ \binom{12 + 3 - 1}{3 - 1} = \binom{14}{2} \]

### Step 5: Calculate \( \binom{14}{2} \)
Using the combination formula:
\[ \binom{14}{2} = \frac{14!}{2!(14-2)!} = \frac{14 \times 13}{2 \times 1} = \frac{182}{2} = 91 \]

### Conclusion
Thus, the total number of ways to keep 12 identical bouquets in 3 identical crates is **91**.

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