$35 \cdot 30$ Label 1 record with 1, 2 records with 2, 3 records with 3, ..., 50 records with 50, and then put these $1+2+3+\cdots+50=1275$ records into a box. If each time one record is taken out and not put back, to ensure that 10 records with the same number are taken out, the minimum number of records that should be taken out is
(A) 10.
(B) 51.
(C) 415.
(D) 451.
(E) 501
(45th American High School Mathematics Examination, 1994)
［Solution］Consider the extreme (worst-case) scenario:
Take out 1 record labeled with No. 1, 2 records labeled with No. 2, $\cdots$, 8 records labeled with No. 8, and 9 records for each of the remaining labels from No. 9 to No. 50, totaling:
$$
1+2+3+\cdots+8+9 \times(50-8)=414,
$$

If one more record is taken, then it can be guaranteed that there will be at least one label with no fewer than 10 records.
Therefore, the answer is $(C)$.