Problem 862

Larger Digit Permutation

For a positive integer $n$ define $T(n)$ to be the number of strictly larger integers which can be formed by permuting the digits of $n$.

Leading zeros are not allowed and so for $n = 2302$ the total list of permutations would be:
$$2023,2032,2203,2230,\mathbf{2302},2320,3022,3202,3220$$
giving $T(2302)=4$.

Further define $S(k)$ to be the sum of $T(n)$ for all $k$-digit numbers $n$. You are given $S(3) = 1701$.

Find $S(12)$.

更大的数字重排

对于正整数$n$，定义$T(n)$为对$n$的数字进行重排所能得到的严格大于$n$的整数数量。

重排不允许有前导零，因此对于$n = 2302$，所有的合法重排包括：
$$2023,2032,2203,2230,\mathbf{2302},2320,3022,3202,3220$$
因此可得$T(2302)=4$。

再定义$S(k)$为所有的$k$位整数$n$对应的$T(n)$之和。已知$S(3) = 1701$。

求$S(12)$。