**A school dance committee is made up of 4 freshman, 2 sophomores, 4 juniors, and 2 seniors.**

**How many ways are there to sit the committee in a row at a meeting if the freshman,sophomores, and juniors must sit by grade, but the seniors can sit wherever they want?**

To solve this problem, we need to calculate the number of ways to seat the committee members under the specified conditions.

### Key Points:
- **Freshmen, sophomores, and juniors must sit by grade**, meaning all members of the same grade must sit together.
- **Seniors can sit anywhere**, without restriction.

#### Step 1: Calculate the arrangements within each group.
- There are 4 freshmen, so the freshmen group can be seated in \(4!\) ways (since all four must be together).
  
  \[
  4! = 4 \times 3 \times 2 \times 1 = 24
  \]

- There are 2 sophomores, so the sophomore group can be seated in \(2!\) ways.
  
  \[
  2! = 2 \times 1 = 2
  \]

- There are 4 juniors, so the junior group can be seated in \(4!\) ways.
  
  \[
  4! = 24
  \]

#### Step 2: Consider the groups as "blocks."
Since the freshmen, sophomores, and juniors must sit together by grade, we treat each grade as a block or "unit." These three groups (blocks) can be arranged among themselves in \(3!\) ways.

\[
3! = 3 \times 2 \times 1 = 6
\]

#### Step 3: Incorporate the seniors.
The 2 seniors can sit anywhere, so they can be seated in \(2!\) ways among themselves.
  
\[
2! = 2
\]

#### Step 4: Arrange the groups and the seniors.
We now have 5 "positions" to fill: 3 grade groups (freshmen, sophomores, and juniors) and 2 individual seniors. We can choose 2 positions for the seniors from these 5 positions. The number of ways to choose 2 positions for the seniors from 5 positions is given by \( \binom{5}{2} \):

\[
\binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10
\]

#### Step 5: Multiply the results.
Now, we combine all the possibilities. The total number of seating arrangements is the product of:
- Arrangements within the freshman, sophomore, and junior groups: \( 4! \times 2! \times 4! = 24 \times 2 \times 24 = 1152 \)
- Ways to arrange the 3 groups: \(3! = 6\)
- Ways to place the seniors: \(\binom{5}{2} = 10\)
- Ways to arrange the seniors: \(2!\)

Thus, the total number of ways to arrange the committee members is:

\[
1152 \times 6 \times 10 \times 2 = 138,240
\]

### Final Answer:
There are **138,240** ways to seat the committee members in a row under the given conditions.

---

Would you like a deeper explanation of any part of the solution? Here are some related questions to further explore:

1. What if the seniors also had to sit together as a group?
2. How would the problem change if there were more seniors?
3. What if only the freshmen had to sit together?
4. How would the result change if the seats were arranged in a circle?
5. How many ways can we seat the committee if all members must sit by grade?

**Tip:** When dealing with arrangement problems, breaking it down into smaller, manageable steps is key to finding the correct solution.

A school dance committee is made up of 4 freshman, 2 sophomores, 4 juniors, and 2 seniors. How many ways are there to sit the committee in a row at a meeting if the freshman, sophomores, and juniors must sit by grade, but the seniors can sit wherever they want?

Discover how to calculate the seating arrangement for a school dance committee with freshmen, sophomores, juniors sitting together by grade, and seniors seated anywhere. This combinatorics problem uses permutations, factorials, and binomial coefficients to find the total number of seating arrangements: 138,240 ways.

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