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Can you create a formula for the amount of possible permutations of a three digit number, who has a digit sum equal to n?

  • (1) When you say permutations, is 1+2+3 different from 1+3 +2 ? (2) Are 0's allowed in the number ? Also allowed at start ? (3) What work have you so far done on it ? – true blue anil Apr 11 '16 at 09:20

2 Answers2

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Since you are looking at 3 digit numbers, the first digit needs to be greater than $0$. This would mean that for the first digit you have $4$ options, $1,2,3,4$. If you start with $1$, then the rest of your digits need to sum to $3$, which would mean that there are $3+1=4$ options available for the second digit (including $0$). For the last digit you have only one option left, which would be $3-second digit$. Thus, starting with $1$ you have a total of $4$ options.

Starting with $2$ you have similarly $2+1$ options left for the second digit, and the last digit is already determined by the first two digits. Hence, you have $3$ options from here.

Starting with $3$ you have $1+1$ options for the second digit, hence, in total. Thus, you have $2$ options from here.

Starting with $4$ you only have $1$ option for the second and third digit, which are zero.

Therefore, you have a total of $4+3+2+1=10$ such permutations. This should give you an idea about generalizing this type of questions. For the three digit case with sum equal to $n\leq 9$, it appears that you have $$ \sum_{i=1}^n i $$ as the formula.

Thomas
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Its stars and bars if $x+y+z=n$ so total ways are ${n+3-1\choose 3-1}={n+2\choose 2}$ but heres a problem it may be possible that $x=0$ and $y=..,z=..$ is a solution but its not a three digit number hence we keep $x=1+l$ so now $1+l+y+z=n$ so now total solutions become ${n+3\choose 2}$ and then get the answer.

Archis Welankar
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