In fact, it is to find a positive integer solution that satisfies the equation, and how many groups are there.
Let a≥b≥c, then1/a ≤1/b ≤1/c, so 3/a ≤1/a+1/c ≤ 3.
When c= 1, the equation becomes1/a+1/b = 3/4-1=-kloc-0//4, which does not meet the conditions.
When c=2, the equation becomes1/a+1/b = 3/4-1/2 =1/4, 1/a≤ 1/b, and use the above method. So in this case, there are three solutions.
When c=3, the equation becomes1/a+1/b = 3/4-1/3 = 5/12, and 1/a≤ 1/b is obtained. So in this case, there are two solutions.
When c=4, the equation becomes1/a+1/b = 3/4-1/4 =1/2, and 1/a≤ 1/b can be obtained.
To sum up, under the condition that a≥b≥c, there are six solutions, namely (20, 5, 2), (12, 6, 2), (8, 8, 2), (12, 2) and (6, 4, 3).
But the size of A, B and C is not clear in the title, so
(20,5,2) has 3x2x 1=6 sequences.
(12,6,2) has 3x2x 1=6 sequences.
(8,8,2) has the order of (3x2x 1)/2=3.
(12,2) There are (3x2x 1)/2=3 orders.
(6,4,3) has 3x2x 1=6 sequences.
(4, 4, 4) There is only one order.
So there are 6+6+3+3+6+ 1=25 solutions, and the answer is B.