For example,the sum of the digits of the number $139=1+3+9=13$The smallest $3$ digit number to fulfill this condition is $109$
Essentially,this question is asking "For three whole numbers $x,y,z$ find the number of permutations such that $x+ y+ z$ $\ge$ 10, $ x \neq 0$(since hundreds digit can't be $0$)"
I want to know how we would calculate the final answer.
I also want to know the formula for this question if it was extended to n digit numbers.
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$\begingroup$Well, There are $900$ three digit numbers. So if we find out how many $3$ digit number have the digits add up to $9$ or less we just subtract those.
Now consider this:
Suppose the three digit number is $abc$.
Suppose you are given $9$ start and $3$ bars and you want to represent you number this way.
From left to right: Put down $a$ stars to represent the first digit. But down $1$ bar to represent a place holder. Put down $b$ more stars to represent the second digit. Put down a second bar to represent a $2$nd place holder. Put down $c$ stars. Put a bar as the $3$rd place holder. You have $a+b+c \le 9$ so you may have some stars remaining. Put them down.
Every three digit number where the sum is $9$ or less can be represented by a unique combination of stars and bars and each combination represents a unique three digit number where the sums of the digits is $9$ or less.
THis is a line of $12$ items and you must choose $3$ positions for where the bars go.
So there are ${12\choose 3}$ such numbers.
But notice we can't have the first digit be $0$. That is, we can't start with a bar.
So given we start with a star we have $11$ more items and we must choose $3$ position for where the bars go.
So tere ${11 \choose 3}$ such number that have three digits, the first digit is not zero, and the digits add to $9$ or less.
SO there are $900-{11\choose 3}$ three digit numbers where the digits add up to $10$ or more.
$\endgroup$ 0 $\begingroup$We could try to evaluate directly how Manu such number exist by conditionning on a particular digit. I'll do it with the last numbers.
If the last digit is $9$. Since the first digit is at least one, any three digits numbers is valid.$$9\times10=90$$If the last digit is $8$. If the first digit is $1$, there is $9$ possibilities for the second digit. Any other value of the first digit have $10$ possibilities for second digit. So$$9+8\times10=89$$If the last digit is $7$. If the first digit is $1$, there is $8$ possibilities for the second digit. If the first digit is $2$, there is $9$ possibilities for the second digit. Any other value of the first digit have $10$ possibilities for second digit. So$$8+9+7\times10=87$$So on for the other cases$$7+8+9+6\times10=84$$$$6+7+8+9+5\times10=80$$$$5+6+7+8+9+4\times10=75$$$$4+5+6+7+8+9+3\times10=69$$$$3+4+5+6+7+8+9+2\times10=62$$$$2+3+4+5+6+7+8+9+1\times10=54$$$$1+2+3+4+5+6+7+8+9+0\times10=45$$All is left is to add those possibilities.$$90+89+87+84+80+75+69+62+54+45=735$$It is longer than @fleablood answer, but it confirm the result since$$900-{11\choose3}=900-165=735$$
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