Suppose that a group of 27 students are going to pose for a class photo.
They are to be arranged in three rows with 5 in the front row, 12 in the middle row and 10 in the back row.
The photographer selects which students go in each row.
In how many different ways can the class now be arranged for the photo?
You can enter as a number, or you can include factorials in your answer, for example you can enter in 5! for 5 factorial.
To multiply a number of terms together you must use *, for example 2!*3!

We need to determine how many different ways the class can be arranged for the photo. The total number of students is 27, and they are to be arranged into three rows: 5 in the front row, 12 in the middle row, and 10 in the back row.

### Step-by-step Approach:

1. **Select 5 students for the front row out of 27:**
   The number of ways to choose 5 students from 27 is given by:

   \[
   \text{Ways to choose 5 students} = \binom{27}{5}
   \]

2. **Select 12 students for the middle row out of the remaining 22:**
   After selecting the 5 students for the front row, we have 22 students left. The number of ways to choose 12 students from 22 is given by:

   \[
   \text{Ways to choose 12 students} = \binom{22}{12}
   \]

3. **The remaining 10 students automatically go to the back row:**
   After selecting 5 students for the front row and 12 for the middle row, the remaining 10 students automatically go to the back row. There is only 1 way to do this.

### Formula:
The total number of ways to arrange the students is:

\[
\text{Total arrangements} = \binom{27}{5} \times \binom{22}{12} \times \binom{10}{10}
\]

Since \(\binom{10}{10} = 1\), it simplifies to:

\[
\text{Total arrangements} = \binom{27}{5} \times \binom{22}{12}
\]

### Factorial Expression:
Recall that the binomial coefficient is expressed as:

\[
\binom{n}{r} = \frac{n!}{r!(n-r)!}
\]

So the formula becomes:

\[
\text{Total arrangements} = \frac{27!}{5! \times 22!} \times \frac{22!}{12! \times 10!} = \frac{27!}{5! \times 12! \times 10!} 
\]

Thus, the final answer is:

\[
\boxed{\frac{27!}{5! \times 12! \times 10!}}
\]

Would you like further details or explanations on any step?

### Related Questions:
1. How would the problem change if there were 6 students in the front row instead of 5?
2. What if the arrangement was different, such as having 8 students in the middle row and 9 in the back row?
3. How can you express the number of ways using permutations instead of combinations?
4. How does the answer change if students in each row need to be ordered as well?
5. Can you derive the formula for a similar arrangement with different numbers of students?

### Tip:
Always break down combinatorial problems into step-by-step selections to avoid errors in calculation.

Discover how to calculate the number of ways to arrange 27 students into three rows (5 in the front, 12 in the middle, 10 in the back) for a class photo. Learn about combinatorial mathematics, binomial coefficients, and factorial calculations. Suitable for Grades 10-12 students.

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