Painful list

integers with one non-zero digit.

09,90 -- 2!

integers with two non-zero digits.

018,081,108,801,810 ---3!

027,...,....,....,.....,......,.----3!

036.............................---3!

045..............................----3!

Sum=4*3!

integers with 3 non-zero digits.

0126,..............................----4!

0135,.................................-----4!

0234,.................................-----4!

sum= 3 * 4!

No integers with 4 non-zero digits because 1+2+3+4>9

Ans 2!+4 3!+3 4!=98

btw: each time i do i get different answers(first 18, then 32, then 86, then this, such a painful list)

If you write out the different possibilities for how this can be achieved with distinct nonzero digits, you will find that there are $27$ ways to do it, put into three sets: $1$ with $1$ digit, $8$ with $2$ digits, and $18$ with $3$ digits. All other possible numbers that fit the problem's conditions have exactly one $0$ somewhere in them, which does not add to the sum of the digits.

For a number with $n$ digits, there are $n$ ways to add a $0$ into the number and create a new number. Counting the number with no $0$ 's in it, there are $n+1$ numbers that can be created for each grouping of digits (for example, the group $54$ can produce $54$ , $540$ , and $504$ ).

To find the answer, multiply the number of digits in each set plus $1$ by the number of elements in that set. $2\times1+3\times8+4\times18=2+24+72=\boxed{98}$

$\textit{Note:}$ Another way to look at this is that you can add the extra $0$ at the front of the number, which retains the original number. This would be saying that the possible numbers created from $54$ are $054=54$ , $504$ , and $540$ .

This seems like a hassle, but I simply listed the possible combinations of digits starting with the largest number in the combination. They are: (9, 0) (8, 1, 0) (7, 2, 0) (6, 3, 0) (6, 1, 2, 0) (5, 4, 0) (5, 3, 1, 0) (4, 3, 2, 0)

Now it remains to find the number of ways to order the numbers (a 0 in front means it doesn't show up in the number) for each possibility and add them up - with factorials. 2+6+6+6+24+6+24+24= 98

9=9 (tier 1) =1+8=2+7=3+6=4+5 (tier 2) =1+2+6=1+3+5=2+3+4 (tier 3) 1 digit numbers: 1 2 digit numbers: (2!)(4)+(1/2)(2!)(1)=9 (arrangements of tier 2 + arrangements of tier 1+0) 3 digit numbers: (3!)(2/3)(4)+(3!)(3)=34 (arrangements of tier 2+0 + arrangements of tier 3) 4 digit numbers: (4!)(3/4)(3)=54 (arrangements of tier 3+0) There cannot be a 5 digit number since 0+1+2+3+4=10>9 Hence total is 98

Honestly when I first saw this question, I thought of using the Stars and Bars technique. However, it doesn't work on this question because of the requirement of having distinct digits. Note that in my following attempt to explain my solution to the question, the words in [ ] are my additional comments.

First, let's go through and narrow down some of the requirements of the question. The question wants us to find how many positive integers with distinct digits that add up to $9$ . Note that negative digits are eliminated here. Also, notice that using non-zero digits, at most 3 distinct digits can create a sum of $9$ . Any more than $3$ distinct non-zero digits add together will create a sum more than $9$ , i.e. $1+2+3+4 > 9$ . Hence, we only have to consider integers with $3$ or less non-zero digits. [Here, I use non-zero integers to simplify the calculation of the problem as you will see below :) ].

Second, I tried to represent the question in a simpler mathematical model. Here, I realised that we can represent the condition of having distinct digits as having each digit greater than the others, i.e. In a $3$ digit non-zero number $\overline{abc}, a>b>c>0$ . From there, the idea now is to simplify this question even more, and to do so, I attempted to convert the $>0$ requirement into $\geq 0$ . Notice that if you minus $1$ from $a$ , minus $2$ from $b$ , and minus $3$ from $c$ in the previous example of $\overline{abc}, a>b>c>0$ , we will get $a-3, b-2, c-1 \geq 0$ where $a-3 + b-2 + c-1 = 9-6 = 3$ . From here, by finding the number of ways of unordered $a,b,c$ that satisfies the condition, then permutate the digits (since the digits can be in any order, e.g. $\overline{bac}$ ), we can obtain the number of non-zero $3$ digit number. [This is just an example of the method that I tried to solve the question: the following will have the explicit cases I have used].

Thirdly, we can observe that integers that satisfy the conditions have either $4$ digits ( $1$ of which is $0$ ), $3$ digits, $2$ digits or $1$ digit. Here we split the question into $3$ (quite weirdly split cases as we are using non-zero digits in our calculation) cases:

Case 1 Integers (positive) with $3$ non-zero digits can be either $4$ digits ( $1$ of which is $0$ ) or $3$ digits with all digits being non-zero. This can be shown as (mentioned earlier): $a-3, b-2, c-1 \geq 0$ and $a-3 + b-2 + c-1 = 9-6 = 3$

See that there are 3 unordered sets of ( $(a-3, b-2, c-1$ ) that satisfy the above, namely $(3,0,0), (2,1,0), (1,1,1)$ . Note that here, $a-3, b-2, c-1$ are not necessarily distinct since after adding back the $1,2$ and $3$ , they will definitely become distinct. Then, by allowing $0642 = 642$ , note that adding a zero in now to form $4$ digit numbers ( $1$ of which is $0$ )or $3$ digits with all digits being non-zero is just multiplying the $3$ possible unordered sets by $4!$ [permutating of the 4 digits including 1 $0$ ]. Hence, $3 \times 4! = 72$

Case 2 Integers with $3$ digits ( $1$ of which is $0$ ) or $2$ digits with all digits being non-zero.

Similarly, we can represent this as $a-2,b-1 \geq 0, a-2 +b-1 = 6$ .

Noting that there are $4$ unordered $(a-2,b-1)$ that satisfy the above, number of solutions to case 2 is $4 \times 3! = 24$ .

Case 3 Integers with $2$ digits ( $1$ of which is $0$ ) or $1$ digit with all digits being non-zero.

We can represent this as $b-1 \geq 0, b-1 = 8$ .

There is only 1 unordered $b-1$ that satisfies the above. Hence the number of solutions to Case 3 is $1 \times 2! = 2$ .

The Last Step We arrived at the answer by simply summing up the number of solutions to the 3 cases:

$72 + 24 + 1 = \boxed{98}$

P.S. Sorry for this looongg solution. I tried to include some thought processes in my solution.

assume set of integers from 0 to 9.

There are 3 ways to grab a set of 4 integers from above that add up to 9: 0126, 0135, and 0234.

Each of these can be permuted 24 ways for a total of 72. We will count 0126 as a 4 digit number even though it is usually represented as the 3 digit number 126.

To that sum of 72, we need to add the permutations of the missing 3 digit numbers that add to 9: 018, 027, 036, and 045.

These permute 6 ways each, for a new grand total of 96. We do NOT need to add the 3 digit sets of 126, 135, and 234 as those are already coverd in the 4 digit set.

Lastly, we need to add the 2 digit set: 09, which permutes 2 ways for a final sum of 98.