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The problem asks:

**"In how many ways can 6 members of a family be assigned to 5 rooms in a hotel such that no room is empty and all 6 members are accommodated?"**

### Understanding the problem:
- There are **6 family members** to be accommodated in **5 rooms**.
- The condition "no room is empty" suggests that each room must have at least one person.
- This problem is essentially a **distribution problem** in combinatorics where we need to distribute 6 people among 5 rooms, ensuring that no room is left empty.

### Solution:

1. **Assigning people to rooms**:
   - The first step is to figure out how to distribute 6 people into 5 rooms such that no room is empty. This is known as a **"surjection problem"** or **"onto function"**.
   - The solution to this can be obtained using **Stirling numbers of the second kind** and multiplying by the number of ways to permute the members within the rooms.

2. **Stirling numbers of the second kind**:
   - The Stirling number \( S(n, k) \) represents the number of ways to partition \( n \) objects into \( k \) non-empty sets. Here, \( S(6, 5) \) represents the number of ways to distribute 6 people into 5 rooms, ensuring no room is empty.
   - \( S(6, 5) = 6! / (1! \cdot 5!) \), which represents the ways to choose 5 rooms from 6 with the restriction.

Let’s calculate \( S(6, 5) \

In how many ways can 6 members of a family be assigned to 5 rooms in a hotel such that no room is empty and all 6 members are accommodated?

Learn how to calculate the number of ways to assign 6 family members to 5 rooms in a hotel such that no room is left empty. This problem explores combinatorics using Stirling numbers of the second kind and permutations. Ideal for students in Grades 10-12.

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