8. There are eight envelopes numbered 1 to 8 . Find the number of ways in which 4 identical red buttons and 4 identical blue buttons can be put in the envelopes such that each envelope contains exactly one button, and the sum of the numbers on the envelopes containing the red buttons is more than the sum of the numbers on the envelopes containing the blue buttons.
(A) 35
(B) 34
(C) 32
(D) 31
(E) 62
8. Answer: (D)
There are $\binom{8}{4}=70$ ways of putting 4 identical red buttons and 4 identical blue buttons in the envelopes. Since $1+2+3+\cdots+8=36$, there are 8 cases where the sum of the numbers on the envelopes containing the red buttons is equal to 18 (which is also equal to the sum of the numbers on the envelopes containing the blue buttons), namely, $(8,7,2,1),(8,6,3,1),(8,5,4,1),(8,5,3,2),(7,6,4,1)$, $(7,6,3,2),(7,5,4,2)$ and $(6,5,4,3)$. Hence it follows that the required number of ways is $\frac{70-8}{2}=31$.