How many descending order number between 100 and 600? Example of valid number are:
(1) 431
(2) 543
(3) 210
(4) 321
Example of not valid number are:
(1) 221
(2) 500
(3) 654
The answer is 120 but I don't know why.
My attempt:
I have to construct a string of length $n=3$ with some constraints.
I split the problem into 5 small problems.
$D_1 = \{\text{all valid number staring with 1}\}$
$|D_1| = 0$
$D_2 = \{\text{all valid number staring with 2}\}$
$|D_2| = 1 \times {2\choose 2}$, because I can choose only from $\{0,1\}$
$D_3 = \{\text{all valid number staring with 3}\}$
$|D_3| = 1 \times {3\choose 2}$, because I can choose only from $\{0,1,2\}$
$D_4 = \{\text{all valid number staring with 4}\}$
$|D_4| = 1 \times {4\choose 2}$, because I can choose only from $\{0,1,2,3\}$
$D_5 = \{\text{all valid number staring with 5}\}$
$|D_5| = 1 \times {5\choose 2}$, because I can choose only from $\{0,1,2,3,4\}$
My solution: $$\sum_{i \in \{1,2,3,4,5\}}|D_i| = 35$$ I can sum directly because there is no double counting but it's wrong...
Editing from the future:
I made a mistake on the summation, the solution is 20. And from the textbook 120 were not the correct answer...
$$\sum_{i \in \{1,2,3,4,5\}}|D_i| = 20$$ I can sum directly because each set is disjoint (there is no double counting). In conclusion my long not-necessary aproach was right but (if you are reading this post) I recommend you to read all the comments below and the answers because they helped me a lot.
 A: Summarizing information from the comments,
a decreasing number of three digits requires three different digits from the set
$\{0,1,2,3,4,5,6,7,8,9\}.$
Once you have chosen the three digits they must appear in the number in decreasing order, so from any one subset of three digits there is only one way to form a decreasing number. So we get a one-to-one correspondence between the three-element subsets of a set of ten elements and the three-digit decreasing numbers.
There are
$$ \binom{10}{3} = 120 $$
three-digit decreasing numbers.
If you put the restriction that the number must be between $100$ and $600,$
the digits must be selected from the six-element set
$\{0,1,2,3,4,5\},$
so there are only
$$ \binom{6}{3} = 20 $$
three-digit decreasing numbers less than $600.$
Note that in the calculation for numbers less than $600,$ we have
$$ \binom 63 = \frac{6!}{3!3!} = \frac{6\cdot 5\cdot 4}{3!} = \frac{120}{6},$$
so if you work out the numerator but forget to divide by $6$
you could mistakenly say the answer is $120$
(which is the correct answer for the number of permutations
of three digits out of six).
A: Another pattern I have recognised is, if we take the first digit as $n$, we have $\frac{(n)(n-1)}{2}$ possibilities for that particular value of $n$. Example: if n=$4$, the possible numbers are {$432,431,430,421,420,410$} which is six in number. Which is also equal to $\frac{4\times3}{2}$.
The reason this works is, suppose if n=$4$, we have three numbers having the second digit as $3$,  two numbers having the second digit as $2$ and one number having the second digit as $1$. Similarly, this works for other values of $n$ too.
so we can say that the total possibilities are: $$\frac{1}{2}\sum_{n=1}^{5}(n)(n-1)$$
Which turns out to be $20$
