of the numbers above.
13. There are 100 students at a school and three dormitories, A, B, and C, with capacities 25,
35, and 40 respectively.
(a). How many ways are there to fill the dormitories?
(b). Suppose that of the 100 students, 50 are men and 50 are women and that A is an all-men’s
dorm, B is an all women’s dorm, and C is co-ed. How many ways are the

Let's break down the two parts of the problem:

### (a) **How many ways are there to fill the dormitories?**

We have 100 students and three dormitories with capacities:
- Dormitory A: 25 students
- Dormitory B: 35 students
- Dormitory C: 40 students

The task is to determine how many ways we can assign the 100 students to these dormitories, respecting their capacities.

This is a combinatorics problem where we are dividing 100 students into three groups with specific sizes. The number of ways to do this can be calculated using the **multinomial coefficient** formula:

\[
\text{Number of ways} = \frac{100!}{25!35!40!}
\]

This represents the number of ways to first select 25 students for dormitory A, then 35 students for dormitory B, and the remaining 40 students for dormitory C.

### (b) **Suppose that of the 100 students, 50 are men and 50 are women and that:**
- Dormitory A is an all-men’s dorm (25 men),
- Dormitory B is an all-women’s dorm (35 women),
- Dormitory C is co-ed (25 men and 15 women).

How many ways are there to assign the students to the dormitories under these restrictions?

First, we need to assign the men and women separately according to the dormitory rules.

#### Step 1: Assign 25 men to Dormitory A
The number of ways to choose 25 men out of 50 is:

\[
\binom{50}{25} = \frac{50!}{25!25!}
\]

#### Step 2: Assign 35 women to Dormitory B
The number of ways to choose 35 women out of 50 is:

\[
\binom{50}{35} = \frac{50!}{35!15!}
\]

#### Step 3: Assign the remaining men and women to Dormitory C
Once we've assigned the men and women to Dormitory A and B, there will be 25 men and 15 women left. All of these remaining men and women will go to Dormitory C.

So, the total number of ways to assign the students to the dormitories is the product of the results from Step 1 and Step 2:

\[
\text{Total ways} = \binom{50}{25} \times \binom{50}{35}
\]

### Summary:
- **(a)** The number of ways to fill the dormitories is \(\frac{100!}{25!35!40!}\).
- **(b)** The number of ways to assign the students under the gender restriction is \(\binom{50}{25} \times \binom{50}{35}\).

Let me know if you'd like me to compute these values or explain any part in more detail.

---

Would you like further details on this solution, or should I compute the exact values for you?

Here are 5 related questions:
1. How do multinomial coefficients work in combinatorics?
2. What is the difference between combinations and permutations?
3. Can you explain the binomial coefficient in more detail?
4. How can we calculate large factorials efficiently?
5. How does the concept of partitions apply to this type of problem?

**Tip:** When dealing with large numbers, breaking the problem into smaller combinatorial steps can help make the calculations more manageable.

There are 100 students at a school and three dormitories, A, B, and C, with capacities 25, 35, and 40 respectively.
(a). How many ways are there to fill the dormitories?
(b). Suppose that of the 100 students, 50 are men and 50 are women and that A is an all-men’s dorm, B is an all women’s dorm, and C is co-ed. How many ways are there to assign the students to the dormitories under these restrictions?

Explore this combinatorics problem involving the assignment of 100 students to three dormitories with specific capacities. Learn how to calculate the number of ways to fill dormitories using multinomial and binomial coefficients. This problem is suitable for undergraduate students studying mathematics or discrete mathematics.

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