This exercise can be solved using combinations even though the problem statement will not always include a form of the word "combination" or "subset."
An electronics store receives a shipment of 30 graphing calculators, including 8 that are defective. Four of the calculators are selected to be sent to a local high school. How many of these selections will contain no defective calculators?
The number of ways to choose all selections that contain no defective calculators is $\square$.
Answer
Final Answer: The number of ways to choose all selections that contain no defective calculators is \(\boxed{7315}\).
Step 1 ：This exercise can be solved using combinations even though the problem statement will not always include a form of the word "combination" or "subset."
Step 2 ：An electronics store receives a shipment of 30 graphing calculators, including 8 that are defective. Four of the calculators are selected to be sent to a local high school. How many of these selections will contain no defective calculators?
Step 3 ：The problem is asking for the number of ways to choose 4 calculators from the 22 non-defective ones. This is a combination problem, because the order in which we choose the calculators does not matter.
Step 4 ：The formula for combinations is: \[C(n, k) = \frac{n!}{k!(n-k)!}\] where n is the total number of items, k is the number of items to choose, and "!" denotes factorial. In this case, n = 22 (the number of non-defective calculators) and k = 4 (the number of calculators to choose).
Step 5 ：Using the formula, we find that the number of combinations is 7315.
Step 6 ：Final Answer: The number of ways to choose all selections that contain no defective calculators is \(\boxed{7315}\).
