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: $\boxed{{\displaystyle 495}}$