$11 \cdot 5$ In the expansion of $(a+b)^{n}$, there are $n+1$ different terms. In the expansion of $(a+b+c)^{10}$, the number of different terms is 
(A) 11.
(B) 33.
(C) 55.
(D) 66.
(E) 132.
(9th American High School Mathematics Examination, 1958)
[Solution] From $(a+b+c)^{10}$
$$
\begin{array}{l}
=[(a+b)+c]^{10} \\
=(a+b)^{10}+10(a+b)^{9} c+\cdots+10(a+b) c^{9}+c^{10}
\end{array}
$$

There are 11 terms in total. However, 
$(a+b)^{10}$ has 11 terms, $(a+b)^{9}$ has 10 terms, $\cdots,(a+b)^{0}$ has 1 term. Therefore, the total number of terms is $1+2+\cdots+10+11=66$. Hence, the answer is $(D)$.