21. 576
21. Let $n$ be such an integer. Note that $n$ is divisible by 4950 if and only if it is divisible by each of 50, 9 and 11. As no two digits of $n$ are the same, the last two digits of $n$ must be 50. Note also that $n$ misses exactly one digit from 0 to 9 with every other digit occurring exactly once. As $n$ is divisible by 9 and the unit digit of $n$ is 0 , it is easy to check that the digit missing must be 9 , and hence the first 7 digits of $n$ must be some permutation of $1,2,3,4,6,7,8$.
Let $n=\overline{A B C D E F G 50}$. Since $n$ is divisible by 11 , we have $A+C+E+G \equiv B+D+F+5(\bmod$ 11). Also, the sum of $A$ to $G$ is $1+2+3+4+6+7+8=31$. It is easy to check that the only possibility is $A+C+E+G=18$ and $B+D+F=13$.
Now there are 4 ways to choose 3 digits from $\{1,2,3,4,6,7,8\}$ to make up a sum of 13 , namely, $\{8,4,1\},\{8,3,2\},\{7,4,2\}$ and $\{6,4,3\}$. Each of these gives rise to 3 ! choices for the ordered triple $(B, D, F)$, and another 4 ! choices for the ordered quadruple $(A, C, E, G)$. It follows that the answer is $4 \times 3!\times 4!=576$.
21. How many nine-digit positive integers consist of nine pairwise distinct digits and are divisible by 4950 ?
有多少個九位正整數的九個數字互不相同, 而且可被 4950 整除？