what if 3 students wont make it and there wld be 805 more ways? solve algebraically

Let's solve this new problem algebraically.

### Step 1: Problem Setup
There are \( n \) students in the school quiz club, and \( r \) students are to be selected for a competition. We are told that 3 students will not make the team, so:
\[
r = n - 3
\]

The problem states that there are 805 more ways to choose and assign the students than to just choose them.

### Step 2: Mathematical Expressions
1. **Number of ways to select the team members (without assignment):**
   \[
   \binom{n}{r} = \binom{n}{n-3}
   \]

2. **Number of ways to select and assign the rounds:**
   \[
   \binom{n}{r} \times r! = \binom{n}{n-3} \times (n-3)!
   \]

### Step 3: Set Up the Equation
We are given that:
\[
\binom{n}{n-3} \times (n-3)! = \binom{n}{n-3} + 805
\]

Simplifying:
\[
\binom{n}{n-3} \times ((n-3)! - 1) = 805
\]

### Step 4: Use the Combination Formula
We know that:
\[
\binom{n}{n-3} = \frac{n!}{(n-3)! \times 3!} = \frac{n \times (n-1) \times (n-2)}{6}
\]

Substitute this into the equation:
\[
\frac{n \times (n-1) \times (n-2)}{6} \times ((n-3)! - 1) = 805
\]

Multiply both sides by 6 to clear the fraction:
\[
n \times (n-1) \times (n-2) \times ((n-3)! - 1) = 4830
\]

### Step 5: Solve the Equation
We'll solve this equation by testing integer values for \( n \).

#### Test \( n = 7 \):
\[
7 \times 6 \times 5 = 210
\]
\[
4! - 1 = 24 - 1 = 23
\]
\[
210 \times 23 = 4830
\]

This satisfies the equation, so \( n = 7 \) is the correct answer.

### Final Answer:
The number of students in the club is \( \boxed{7} \).

Do you want further details or have any questions?

Here are some related questions to explore:
1. What happens if 4 students won't make it and the difference in ways is 1260?
2. How does changing the value of \( r \) affect the solution?
3. Can you solve this problem using a different approach, like a direct factorial method?
4. What if the extra ways were not 805 but another value, say 500?
5. How would you approach this problem if the assignment of rounds was random?

**Tip:** When solving algebraic problems that involve factorials, check for integer solutions by substituting small values early on to simplify the process.

Learn how to solve an algebraic problem involving the selection of students for a club competition. Understand the relationship between combinations, factorials, and equations to determine the correct number of students. Suitable for high school students.

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