How many three-digit numbers are there in which all digits on the left are larger than digits on the right?
This was a problem from an online test, which I gave a wrong answer. The correct answer is $120$, but here is how I attempted it:
Attempt #1: Understand the problem as "All three-digit numbers $\overline{abc}$ that satify $a>c$".
$b$ can be any digit from $0$ to $9$ so there will be $10$ possible numbers for each of the following cases:
If $a=1$ then $b=0$ ($1$ case)
If $a=2$ then $b\in{(0;1)}$ ($2$ cases)
If $a=3$ then $b\in{(0;1;2)}$ ($3$ cases)
If $a=4$ then $b\in{(0;1;2;3)}$ ($4$ cases)
...
- If $a=9$ then $b\in{(0;1;2;3;4;5;6;7;8)}$ ($9$ cases)
So the total number of numbers is $\sum_{x=1}^{9}(x)\times 10=450$ numbers.
Attempt #2 was made after seeking the solution: Understand the problem as "All three-digit numbers $\overline{abc}$ that satify $a>b>c$".
If $a=2$ then $b=1$ and $c=0$.
If $a=3$ then: if $b=1$ then $c=0$ ($1$ case); if $b=2$ then $c\in{(0;1)}$ ($2$ cases).
If $a=4$ then: if $b=1$ then $c=0$ ($1$ case); if $b=2$ then $c\in{(0;1)}$ ($2$ cases); if $b=3$ then $c\in{(0;1;2)}$ ($3$ cases)
...
- If $a=9$ then: fo4 $b=1$ to $b=8$ the amount of possible cases are $1;2;3;4;5;6;7;8$ respectively.
So the total number of numbers that I calculated is truly $120$ for this case.
Is my interpretion correct or is the problem unclear?
