Take any four blocks, * * there are C 12, 4 = 495 methods;

The method of excluding all first-class products: C5, 4 = 5, then the method is = 495-5 = 490;

490 is the final answer.

Solution 2:

Not all of them are first-class, including four situations: first-class 0, 1 first-class, first-class 2, first-class 3;

Quantity of 0 first-class products: C5, 0 * C7, 4 = 35;

1 first-class product quantity: C5, 1 * C7, 3 =175;

The number of two first-class products: C5, 2 * C7, 2 = 210;

Number of third-class products: C5, 3 * C7,1= 70;

Then all the methods * * * are: 35+175+210+70 = 490;

490 is the final result.

I hope it helps you _