3. Arrange 4 identical red balls and 4 identical blue balls in a row, with each ball corresponding to a sequence number from $1,2, \cdots, 8$ from left to right. If balls of the same color are indistinguishable, the number of arrangements where the sum of the sequence numbers corresponding to the 4 red balls is less than the sum of the sequence numbers corresponding to the 4 blue balls is ( ).
(A) 31
(B) 27
(C) 54
(D) 62
3. A.
1 to 8, taking any four different numbers to find their sum, we can get $\mathrm{C}_{8}^{4}=70$ answers (which can be the same).
Among them, the sum being 18 has 8 cases:
$$
\begin{array}{l}
(8,7,2,1),(8,6,3,1),(8,5,4,1), \\
(8,5,3,2),(7,6,4,1),(7,6,3,2), \\
(7,5,4,2),(6,5,4,3) .
\end{array}
$$

Therefore, the number of arrangements where the sum of the indices corresponding to 4 red balls is less than the sum of the indices corresponding to 4 blue balls is $\frac{70-8}{2}=31$.