m distinct animals of a circus have to be placed in m cages, one in cach cage. If n (< m) cages are too small to accommodate p (n < p < m) animals, then the number of ways of putting the animals into cages are

Options

A. `(""^(m-n) P_p)^(m-n) P_(m-p)`
B. `""^(m-n)C_(p)`
C. `(""^(m-n)C_(p))(""^(m-p)C_(m-p))`
D. None of these

Answer

To solve the problem of placing m distinct animals into m cages, given that n cages are too small to accommodate p animals, we can break it down into a step-by-step solution.

### Step-by-Step Solution:

1. **Identify the Total Number of Cages and Animals**:
- We have m distinct animals and m cages.
- Among these m cages, n cages are too small to accommodate p animals.

2. **Determine the Usable Cages**:
- Since n cages cannot accommodate p animals, the remaining cages that can accommodate the animals are \( m - n \).

3. **Choose the Animals for the Small Cages**:
- We need to select p animals from the m distinct animals to place in the n small cages.
- The number of ways to choose p animals from m animals is given by the combination formula \( C(m, p) \).

4. **Arrange the Chosen Animals in the Small Cages**:
- Once we have selected p animals, we can arrange these p animals in the n small cages. Since the cages are distinct, the number of arrangements is \( p! \).

5. **Place the Remaining Animals in the Remaining Cages**:
- After placing p animals in the n small cages, we have \( m - p \) animals left.
- These remaining animals can be placed in the remaining \( m - n \) cages.
- The number of ways to arrange these \( m - p \) animals in \( m - n \) cages is \( (m - p)! \).

6. **Combine All the Calculated Values**:
- The total number of ways to place the animals into the cages is given by multiplying the number of ways to choose the animals, arrange them, and arrange the remaining animals:
\[ \text{Total Ways} = C(m, p) \times p! \times (m - p)! \]

7. **Simplify the Expression**:
- Using the property of combinations, \( C(m, p) = \frac{m!}{p!(m-p)!} \), we can simplify:
\[ \text{Total Ways} = \frac{m!}{p!(m-p)!} \times p! \times (m - p)! = m! \]

### Final Answer:
The total number of ways of putting the animals into cages is \( m! \).