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Problem 862


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:
2023,2032,2203,2230,2302,2320,3022,3202,3220
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(一)

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:
2023,2032,2203,2230,2302,2320,3022,3202,3220
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

S(12)


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