How many $5$-digit numbers are there with digits $abcde$ such that $a < b < c \leq d \leq e$? 
How many $5$-digit numbers are there with digits $abcde$ such that $a < b < c \leq d \leq e$?

I know that if all of them were distinct, then the answer will be $C(9,5)$. However, there are some "less than or equal to" notations, so I do not know how to approach it. I need help. Thanks in advance.
 A: Hint: If you replace $d$ by $d+1$, and $e$ by $e+2$ in your number, then you are looking for all distinct digits, and now the digits are at most 11 (instead of at most 9).
A: The first thing to note is that if we select k distinct digits from n digits, the digits can be arranged in a strictly ascending order in only one way
We can break up the number string into two parts
(a): $a,b,c:\;$ The ending digit($c$) can range from $3-9$ and it can be seen that if the ending digit $c$ is $n$, say, from the lower digits, there can be $\binom{n-1}2$ selections in strictly ascending order
(b): Suppose c was $3$, then we have $10-3=7$ digits for the remaining slot $e$ . These slots can either have distinct digits $\binom72$, or identical digits, $7$
Combining the two parts, the formula comes out to
$$\sum_{n=3}^{10}\binom{n-1}2\left[\binom{10-n}2+(10-n)\right] = 462$$
A: To find all digits such that $0<a < b < c \leq d \leq e$,

*

*find all subsets $\{a,b,c,d,e\}$ from $\{1,\ldots,9\}\ $ (w.l.o.g $a<b<c<d<e$) to get all 5 digit numbers, there are $\binom 9   5$ such numbers

*find all subsets $\{a,b,c,e\}$ from $\{1,\ldots,9\}\ $ (w.l.o.g $a<b<c<e$), set $d:=c$ to get all 5 digit numbers such that $a<b<c=d<e$, there are $\binom 9   4$

*find all subsets $\{a,b,c,d\}$ from $\{1,\ldots,9\}\ $ (w.l.o.g $a<b<c<d$), set $e:=d$ to get all 5 digit numbers such that $a<b<c<d=e$, again there are $\binom 9   4$ such numbers

*find all subsets $\{a,b,c\}$ from $\{1,\ldots,9\}\ $ (w.l.o.g $a<b<c$), set $d:=c$, $e:=c$, to get all 5 digit numbers such that $a<b<c=d=e$, there are $\binom 9   3$
So all in all there are $ \binom 9   5 + 2 \binom 9   4 +\binom 9   3 = 462$ such numbers
