For today's discussion, we will look at one of the more recent Daily Magic Spell.

**Problem Statement:**

Suppose you have an $8$-digit number. How many such numbers have the sum of their digits being $10$?

**Official Solution****:**

Suppose that $abcdefgh$ is a $10$ digit number. Then we want $a+b+c+d+e+f+g+h = 10$. Since $a \geq 1$, we must put one ball in the first box, leaving $9$ balls to distribute in $8$ boxes.

$\displaystyle \binom{9+8-1}{9} = 11440$

However, we cannot place all $10$ balls in the first bin since this would indicate that the first digit is $10$. Therefore, the number of $8$ digit numbers with digit sum $10$ is

$11440 - 1 = 11439$

**Many students obtained the correct solution but didn't receive credit. This error should be fixed!**

The solution accounted for the condition when these numbers are palindromes where the question never addressed that. The solution is now fixed!

One most common problem with this question is that most people would initially interpret this question as a number theory problem. Because of this, it may confuse students on how to approach this question.

Another potential mistake that students might come across is the possibility of forgetting to account how the first digit of the $8-$digit number cannot be $0$, because of this, students may develop a similar Stars and Bars approach, but instead of reserving one ball in the first box, they reserve none. Therefore, their answer would be $\displaystyle \binom{10+8-1}{10} = 19448$.

Another common mistake that students could potentially make is that they forget to account for the possibility of putting all $10$ balls in the first bin. Therefore, students may forget to subtract $1$ from the total.

Let us know other possible approaches that you may have in solving this problem! If you have any requests for another Daily Magic Spell problem, please post below for suggestions! Thank you!