Number of $3$-digit numbers with digits in descending order

combinatorics

Question:

How many $3$-digit numbers are possible in which the digits are in descending order?

Now, I did: $\frac{9\times9\times8}6$ while the video did: $\frac{10\times9\times8}6$.

As apparent, the video considered all possible digits that can come in $3$ places and then dividing by $6$ will leave only those $3$-digit numbers with their digits in descending order.
Now I also reasoned exactly the same but naturally, hundred's digit can't have $0$ so didn't include it but in my mental imagery all the numbers were resulting in what was asked by the question. I know I'm wrong. But I can't "see" (mentally) how/why?
Please help.

Note: I'm not really interested in solving the question by some other method though I'll be glad to know them. My main concern is to check and correct my logical fallacy so I can tackle similar problems in the future without repeating the same mistake.

Edit:
Further explanation of the thought process involved into $\frac{9\times9\times8}6$:
The $100$s' digit may contain digits from $1$ to $9$, then $10$s' digit may contain numbers from $0$ to $9$ except the digit now $100$'s place is using. Finally, unit's place may also contain digits form $0$ to $9$ except the digits now $100$s' and $10$s' place are using. Now I multiply them: $9\times9\times8$. Now to select the ones with specific descending order, I divide by $6$.

Best Answer

As the other answers have explained, $\frac{10\cdot9\cdot8}{6}=8\cdot3\cdot5=120$ is the correct answer. To show why your answer is wrong, I'm first going to make an educated guess about how you came to your solution.

For a 3 digit number, there are $9$ choices for the first digit, $9$ choices for the second, and $8$ for the third. Thus there are $9\cdot9\cdot8$ three-digit numbers with unique digits.

Now for every set of 3 numbers, $\color{red}{\text{there are 6 three-digit numbers}}$, only one of which is sorted in decreasing order. Thus the number of descending three-digit numbers is one-sixth of the total amount of three-digit numbers with distinct digits.

Thus there are $\frac{9\cdot9\cdot8}{6}$ three-digit numbers with digits sorted in decreasing order.

Note the expression in red. It is false. This is because, when excluding numbers that start with zero, you've made it so that sets of three numbers which contain zero only appear 4 times among the 3-digit numbers.

With this knowledge, you can correct your solution by tracking the numbers with zero separately.

There are $9\cdot8\cdot7$ numbers with no zeros, $9\cdot1\cdot8$ numbers with a zero in the tens place, and $9\cdot8\cdot1$ numbers with a zero in the ones place. This gives a total of $9\cdot8\cdot2$ numbers with a zero.

From the numbers with no zeros, there are six numbers per set of digits with one decreasing. Thus we get $\frac{9\cdot8\cdot7}{6}=4\cdot3\cdot7$ decreasing numbers from here.

From the numbers with a zero, there are four numbers per set of digits, with one decreasing. Thus we get $\frac{9\cdot8\cdot2}{4}=4\cdot9$ decreasing numbers.

Adding these totals together, we get $(4\cdot3\cdot7)+(4\cdot9)=12\cdot(7+3)=120$ numbers with decreasing digits.

Which is the amount given by the solutions.

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