1, first consider the first two points on a diameter, so that * * * has 2n/2=n cases; Then consider the third point. Because the positions of the two points have been determined, there are 2n-2 cases of the third point, so the number of right triangles is n(2n-2)=2n(n- 1).

2.i. Only one number is 4 or 7: 2× 8× 8+8× 2× 8+8× 8× 2× 8+8× 8× 8× 8 = 4096.

Two. There are two digits that are 4 or 7: 4! /(2! ×2! )×4×8×8= 1536

Three. There are three numbers that are 4 or 7: 4! /(3! × 1)×2^3×8=256

Ⅳ. All numbers are 4 or 7: 2 4 =16.

* * * There are 4096+1536+256+16 = 5904.