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will it be : 15 * 16^(n-4) * (n-1)P3 + 2 *16^(n-3) * (n-1)P2
where (n-1)P3 is like nPr.
first 15: let's say 0, 1 or A is not the most significant digit. so there will be 15 posibilities (except 0) for that digit.
16^(n-4): now within remaining (n-1) digits we will have to fix 0, 1 and A somewhere. which will allow n-4 digits to be selected freely from 16 digits.
(n-1)P3 : we can select any 3 places out of (n-1) digits that can have 0,1 and A. here order of 0,1 and A is important so used permutation.
+ other half of equation:-
first 2: let's say any one of 0,1 and A is first digit (most significant). But first digit can't be 0. So we have two choices for first digit.
16^(n-3) : first digit is fix and other two digits 0 and 1/A need to be fixed. so remaining (n-3) digits can be selected from 16 digits.
(n-1)P2 : first digit (1 or A) is fixed and other 2 digits 0 and (1 or A) need to be placed within (n-1) places. So we can select any 2 places from (n-1) and order is important. so took permutation.
Are my assumptions correct? Am I missing anything?
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Project Euler #162: Hexadecimal numbers
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will it be : 15 * 16^(n-4) * (n-1)P3 + 2 *16^(n-3) * (n-1)P2
where (n-1)P3 is like nPr.
first 15: let's say 0, 1 or A is not the most significant digit. so there will be 15 posibilities (except 0) for that digit.
16^(n-4): now within remaining (n-1) digits we will have to fix 0, 1 and A somewhere. which will allow n-4 digits to be selected freely from 16 digits.
(n-1)P3 : we can select any 3 places out of (n-1) digits that can have 0,1 and A. here order of 0,1 and A is important so used permutation.
+ other half of equation:-
first 2: let's say any one of 0,1 and A is first digit (most significant). But first digit can't be 0. So we have two choices for first digit.
16^(n-3) : first digit is fix and other two digits 0 and 1/A need to be fixed. so remaining (n-3) digits can be selected from 16 digits.
(n-1)P2 : first digit (1 or A) is fixed and other 2 digits 0 and (1 or A) need to be placed within (n-1) places. So we can select any 2 places from (n-1) and order is important. so took permutation.
Are my assumptions correct? Am I missing anything?