$11 \cdot 12$ If $(3 x+1)^{5}=a x^{5}+b x^{4}+c x^{3}+d x^{2}+e x+f$, then the value of $a-b+c-d+e-f$ is
(A) -32.
(B) 32.
(C) 1024.
(D) -1024.
(Xi'an, Shaanxi Province, China Junior High School Mathematics Competition, 1990)
[Solution] When $x=-1$, the right side of the equation is $=-a+b-c+d-e+f$, $\therefore \quad a-b+c-d+e-f=-[3 \cdot(-1)+1]^{5}=32$. Therefore, the answer is $(B)$.