An electronics store receives a shipment of 19 graphing calculators, including 7 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 slections that contain no defective calculators is:
Number

Step 1 ：Translate the problem into English: An electronics store receives a shipment of 19 graphing calculators, including 7 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 2 ：Find the number of ways to choose 4 calculators from the 12 non-defective ones (since there are 19 total calculators and 7 are defective).

Step 3 ：Use the formula for combinations: C(n, k) = n! / (k!(n-k)!) where n = 12 (non-defective calculators) and k = 4 (calculators to be selected).

Step 4 ：Calculate the combinations: \( C(12, 4) = \frac{12!}{4!(12-4)!} \)

Step 5 ：\( C(12, 4) = \frac{12!}{4!8!} \)

Step 6 ：\( C(12, 4) = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} \)

Step 7 ：\( C(12, 4) = \frac{12 \times 11 \times 10 \times 9}{24} \)

Step 8 ：\( C(12, 4) = 495 \)

Step 9 ：Final Answer: \(\boxed{495}\)