Any two positions of (1) 10 can form an ordinal number pair, * * c (10/0,2) =10× 9 ÷ (1× 2) = 45 pairs.

(2) The first five groups and the last five groups are sequential, * * 2c (5 5,2) = 2× (5× 4)/(1x2) = 20.

There are five pairs 1, four pairs of 3, three pairs of 5, two pairs of 7 and one pair of 9 in the front group and the back group, * * * 50 40 30 20 1= 15, a total of 35 pairs.

(3)***45 pairs, set X in sequence and set 45-x in reverse order.

X-(45-x)=2x-45, even and odd, the result is still odd, and there can be no even difference.