A shipment of 9 motorcycle sets contains 4 defective sets. In how many ways can a company purchase 4 of these sets and receive at least 2 of the defective sets?

We are given a shipment of 9 motorcycle sets, 4 of which are defective. The task is to determine in how many ways a company can purchase 4 of these sets and receive at least 2 defective ones.

### Step 1: Define the possible cases
We need to consider how many defective sets are selected out of the 4 defective sets, with the rest being non-defective. Since the company must receive at least 2 defective sets, we have the following possible cases:

1. **Case 1: 2 defective sets and 2 non-defective sets**  
2. **Case 2: 3 defective sets and 1 non-defective set**  
3. **Case 3: 4 defective sets and 0 non-defective sets**

### Step 2: Calculate the number of ways for each case

#### Case 1: 2 defective sets and 2 non-defective sets
- We choose 2 defective sets from the 4 defective sets:  
  \[
  \binom{4}{2} = \frac{4 \times 3}{2 \times 1} = 6
  \]
- We choose 2 non-defective sets from the 5 non-defective sets:  
  \[
  \binom{5}{2} = \frac{5 \times 4}{2 \times 1} = 10
  \]
- Total for this case:  
  \[
  6 \times 10 = 60
  \]

#### Case 2: 3 defective sets and 1 non-defective set
- We choose 3 defective sets from the 4 defective sets:  
  \[
  \binom{4}{3} = 4
  \]
- We choose 1 non-defective set from the 5 non-defective sets:  
  \[
  \binom{5}{1} = 5
  \]
- Total for this case:  
  \[
  4 \times 5 = 20
  \]

#### Case 3: 4 defective sets and 0 non-defective sets
- We choose all 4 defective sets:  
  \[
  \binom{4}{4} = 1
  \]
- We choose 0 non-defective sets from the 5 non-defective sets:  
  \[
  \binom{5}{0} = 1
  \]
- Total for this case:  
  \[
  1 \times 1 = 1
  \]

### Step 3: Sum the results
Now, sum the results from all three cases:
\[
60 + 20 + 1 = 81
\]

### Final Answer:
There are \( 81 \) ways for the company to purchase 4 motorcycle sets and receive at least 2 defective ones.

Would you like more details on any of the steps?

Here are 5 related questions for further exploration:
1. How many ways can the company purchase exactly 1 defective set?
2. What is the probability that the company will receive exactly 3 defective sets if they choose 4 sets randomly?
3. How would the result change if there were 6 defective sets out of 12 total sets?
4. What if the company wants to purchase 5 sets instead of 4? How does that affect the number of combinations?
5. How many ways are there to purchase 4 sets such that none of them are defective?

**Tip:** Remember that \(\binom{n}{k}\), the combination formula, calculates how many ways to choose \(k\) items from \(n\) without regard to order, and is a key tool in probability and combinatorics problems.

Explore the combinatorial problem of choosing 4 motorcycle sets from a shipment of 9, with at least 2 defective sets. This detailed solution involves combinatorial methods, using the combination formula to calculate the number of ways to achieve this scenario. Ideal for students in Grades 9-12.

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