In a circus there are ten cages for accommodating ten animals Out of these four cages are so small that five out of 10 animals cannot enter into them. In how many ways will it is possible to accommodate ten animals in these cages

Answer

To solve the problem of accommodating 10 animals in 10 cages, where 4 of the cages are too small for 5 of the animals, we can break down the solution into several clear steps.

### Step-by-Step Solution:

1. **Identify the Total Animals and Cages**:
We have a total of 10 animals and 10 cages. Out of these cages, 4 are small and cannot accommodate 5 of the animals.

2. **Determine the Animals that Can Fit in Small Cages**:
Since 5 animals cannot enter the small cages, it implies that only 5 animals can be accommodated in the small cages.

3. **Choose Animals for Small Cages**:
We need to select 5 animals from the total of 10 to place in the small cages. The number of ways to choose 5 animals from 10 is given by the combination formula:
\[ \text{Number of ways} = \binom{10}{5} \]

4. **Calculate the Combination**:
\[ \binom{10}{5} = \frac{10!}{5!(10-5)!} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

5. **Arrange the Selected Animals in Small Cages**:
The 5 selected animals can be arranged in the 4 small cages. Since there are 4 cages, we can arrange the 5 animals in these cages. The number of ways to arrange 5 animals in 4 cages is given by:
\[ 4! \text{ (for the cages)} \times 1 \text{ (for the extra animal)} \]
However, since one cage will have to accommodate more than one animal, we can simply use the factorial of the number of animals to arrange them:
\[ 5! = 120 \]

6. **Arrange the Remaining Animals**:
After placing 5 animals in the small cages, we have 5 remaining animals. These 5 animals can be placed in the remaining 6 cages (since 4 cages are already occupied). The number of ways to arrange these 5 animals in 6 cages is:
\[ 6! = 720 \]

7. **Calculate the Total Arrangements**:
The total number of ways to accommodate all 10 animals in the cages is the product of the number of ways to choose the animals for the small cages, arrange them, and arrange the remaining animals:
\[ \text{Total Ways} = \binom{10}{5} \times 5! \times 6! = 252 \times 120 \times 720 \]

8. **Final Calculation**:
\[ 252 \times 120 = 30240 \]
\[ 30240 \times 720 = 21772800 \]

Thus, the total number of ways to accommodate the 10 animals in the cages is **21,772,800**.