Problem 67 - Precalculus with Limits

Question

A shipment of 25 television sets contains three defective units. In how many ways can a vending company purchase four of these units and receive (a) all good units, (b) two good units, and (c) at least two good units?

Step-by-Step Solution

Answer

Therefore, the number of ways for each scenario are: (a) for all good units, calculate \( \binom{22}{4} \), (b) for two good units, calculate \( \binom{22}{2} \times \binom{3}{2} \), (c) for at least two good units, calculate the sum of \( \binom{22}{2} \times \binom{3}{2} \), \( \binom{22}{3} \times \binom{3}{1} \), and \( \binom{22}{4} \)

Step 1: Find Total Number of Ways to Select Four Units

There are 25 units in total and the company wants to purchase four units. The total number of ways to select four units out of 25, not considering whether the units are defective or not, can be found using the combination formula: \( \binom{25}{4} = \frac{25!}{4!(25-4)!} \)

Step 2: Find Number of Ways to Select All Good Units

There are 22 good units (25 - 3 defective units). The number of ways to select all good units can be found by substituting 'n' with 22 (total number of good units) and 'r' with 4 (units to purchase) in the combination formula: \( \binom{22}{4} = \frac{22!}{4!(22-4)!} \)

Step 3: Find Number of Ways to Select Two Good Units

Here the company wants to select two good units and two defective units. The number of ways to select two good units out of 22 can be found using the combination formula: \( \binom{22}{2} = \frac{22!}{2!(22-2)!} \). Similarly, the number of ways to select two defective units out of 3 can be found using the same formula: \( \binom{3}{2} = \frac{3!}{2!(3-2)!} \). As we want two good and two defective units, multiply these two results to get the total.

Step 4: Find Number of Ways to Select At Least Two Good Units

In this scenario, there can be 2, 3, or 4 good units. To compute this, find the number of ways for each of these cases and then sum the results. The number of ways to select 2 good and 2 defective units has been calculated in the previous step. For 3 good and 1 defective, use the similar method: \( \binom{22}{3} \times \binom{3}{1} \), and for 4 good units it has been calculated in the step related to 'all good units'. Sum these three sets of results to get the total.