First I'll cite some sources. This method has been described by Hans Bodlaender, and it is closely related to the quasistandard numbering system devised by Reinhard Scharnagl.
Suppose we place the bishops first. There are 4 squares available to each bishop, and therefore 4*4 = 16 ways to place the pair. Next we place the queen on one of the 6 remaining squares. Then the knights; there are 5*4/2 = 10 ways to place the two knights on the five remaining squares. Finally three squares are left for the king and rooks, and there is only 1 way to place them, since the king must be between the two rooks. Thus there are 16*6*10*1 = 960 possible positions.
The important point is that, in the above counting, the number of placements available to any given piece type is independent of where the preceding pieces were placed. For example, once the two bishops are placed, there are 6*10 = 60 ways to place the remaining pieces, and this is true whether the bishops were placed on a1 and f1, or on d1 and e1, or wherever. Thus, by placing the bishops first, we select one of 16 classes of positions, with the same number of positions in each class. It is therefore 'safe' to place the bishops first.
By contrast, if we place the king first, then the number of possibilities for the remaining pieces depends on where the king is placed. If the king is on b1, then one rook must be on a1, and the other can be anywhere from c1 to h1. Thus with the king on b1 there are 1*6 = 6 ways to place the rooks. But if the king is on c1, there are 2*5 = 10 ways to place the rooks, and if the king is on d1, there are 3*4 = 12 ways to place the rooks. (Also the number of possibilities for the bishops depends on how the preceding pieces are distributed between the two colors of squares.) By placing the king first we select one of 6 classes of positions, but the various classes contain different numbers of positions, and therefore this method skews the probabilities.
Suppose we place the bishops first. There are 4 squares available to each bishop, and therefore 4*4 = 16 ways to place the pair. Next we place the queen on one of the 6 remaining squares. Then the knights; there are 5*4/2 = 10 ways to place the two knights on the five remaining squares. Finally three squares are left for the king and rooks, and there is only 1 way to place them, since the king must be between the two rooks. Thus there are 16*6*10*1 = 960 possible positions.
The important point is that, in the above counting, the number of placements available to any given piece type is independent of where the preceding pieces were placed. For example, once the two bishops are placed, there are 6*10 = 60 ways to place the remaining pieces, and this is true whether the bishops were placed on a1 and f1, or on d1 and e1, or wherever. Thus, by placing the bishops first, we select one of 16 classes of positions, with the same number of positions in each class. It is therefore 'safe' to place the bishops first.
By contrast, if we place the king first, then the number of possibilities for the remaining pieces depends on where the king is placed. If the king is on b1, then one rook must be on a1, and the other can be anywhere from c1 to h1. Thus with the king on b1 there are 1*6 = 6 ways to place the rooks. But if the king is on c1, there are 2*5 = 10 ways to place the rooks, and if the king is on d1, there are 3*4 = 12 ways to place the rooks. (Also the number of possibilities for the bishops depends on how the preceding pieces are distributed between the two colors of squares.) By placing the king first we select one of 6 classes of positions, but the various classes contain different numbers of positions, and therefore this method skews the probabilities.