Answer the questions based on the following information. Hirawala is a uber rich guy who lives in a swanky place, called Hiranandini, in Mumbai. He has 15 diamond rings and 5 daughters.
If the rings are distinct, find the number of ways of distributing these rings among his 5 daughters such that no any two daughters get equal number of rings but each daughter gets at least one ring and at most 5 rings.

Options

A. `^15C_5`
B. `(15!)/(1!2!3!4!5!)xx5`
C. `(15!)/(1!2!3!4!5!)`
D. `(15!)/(1!2!3!4!)`

Answer

To solve the problem of distributing 15 distinct diamond rings among 5 daughters such that no two daughters receive the same number of rings, and each daughter gets at least one ring and at most 5 rings, we can follow these steps:

### Step 1: Determine the Distribution of Rings
Since each daughter must receive a different number of rings, and they can receive between 1 to 5 rings, the only possible distribution that meets these criteria is:
- Daughter 1: 1 ring
- Daughter 2: 2 rings
- Daughter 3: 3 rings
- Daughter 4: 4 rings
- Daughter 5: 5 rings

This distribution uses all 15 rings (1 + 2 + 3 + 4 + 5 = 15).

### Step 2: Calculate the Number of Ways to Distribute the Rings
Since the rings are distinct, we need to calculate the number of ways to assign the rings to each daughter based on the distribution determined in Step 1.

1. **Choose 1 ring for Daughter 1:** There are 15 options.
2. **Choose 2 rings for Daughter 2:** After giving 1 ring to Daughter 1, there are 14 rings left. The number of ways to choose 2 rings from 14 is given by the combination formula \( C(n, k) = \frac{n!}{k!(n-k)!} \). Thus, the number of ways to choose 2 rings from 14 is \( C(14, 2) = \frac{14!}{2!(14-2)!} = \frac{14 \times 13}{2 \times 1} = 91 \).
3. **Choose 3 rings for Daughter 3:** After distributing rings to Daughters 1 and 2, there are 12 rings left. The number of ways to choose 3 rings from 12 is \( C(12, 3) = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 220 \).
4. **Choose 4 rings for Daughter 4:** After distributing rings to Daughters 1, 2, and 3, there are 9 rings left. The number of ways to choose 4 rings from 9 is \( C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126 \).
5. **Assign the remaining 5 rings to Daughter 5:** There are 5 rings left, and Daughter 5 will receive all of them. There is only 1 way to do this.

### Step 3: Calculate the Total Number of Ways
Now, we multiply the number of ways for each daughter:
- For Daughter 1: 15 ways
- For Daughter 2: 91 ways
- For Daughter 3: 220 ways
- For Daughter 4: 126 ways
- For Daughter 5: 1 way

Thus, the total number of ways to distribute the rings is:
\[ 15 \times 91 \times 220 \times 126 \times 1 \]

### Step 4: Calculate the Final Result
Now we compute the product:
1. \( 15 \times 91 = 1365 \)
2. \( 1365 \times 220 = 300300 \)
3. \( 300300 \times 126 = 37837800 \)

Therefore, the total number of ways to distribute the rings is **37,837,800**.

### Final Answer
The total number of ways to distribute the 15 distinct diamond rings among 5 daughters such that no two daughters get the same number of rings is **37,837,800**.

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