How many 3 digit numbers are there such that the sum of its digits $\ge$ 10?

Question

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.

Answer 1

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 $9$ stars 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 there are ${11 \choose 3}$ 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.

Answer 2

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 $$900-{11\choose3}=900-165=735$$