Permutations: Number of numbers possible I want you to point out mistake in my answer to the question given below

A seven digit number is in the form of $abcdefg$ ( g,f,e,etc. are digit at units, tens, hundreds place etc.) 
  where $a<b<c<d>e>f>g$. Then the number of such possible numbers is
1) 1960
2) 4800
3) 7608
4) 4704

This is how I solved:-
For first four digits abcd we select four numbers from set 1 to 9 (as they will automatically be arranged in order) thus $$^9C_4$$ and then for the rest three we will count each step according to value of d, thus if 
$d=4$ we have four choices for e,f,g (as 3,2,1,0) thus $^4C_3$
$d=5$ we have five choices for e,f,g (as 4,3,2,1,0) thus $^5C_3$
similarly going on till d=9, we will get $$^4C_3+^5C_3+.....+^9C_3$$
Thus total numbers = $$(^9C_4)(^4C_3+^5C_3+.....+^9C_3)$$ but my answer is way too large than any of options, what am I doing wrong? 
The answer to the question is option (3).
 A: You have to select the values for $a,b,c$ in a way similar to the values of $e,f,g$. The difference is that $0$ is not possible for $a,b$ or $c$.
So if $d=4$ then we have $^4C_3$ choices for $e,f,g$ but only $^3C_3$ choices for $a,b,c$ and therefore $^4C_3\cdot ^3C_3$ choices for selecting $a,b,c,e,f,g$, because every selection of $a,b,c$ can be combined with every selection of $e,f,g$.
Generally we have $^dC_3\cdot ^{d-1}C_3$ possibilities to select $a,b,c,e,f,g$ if $d$ is given. The value of $d$ can be one of $4,5,6,7,8,9$. 
Therefore we have
$$^3C_3\cdot ^4C_3+^4C_3 \cdot ^5C_3+\cdots +^8C_3 \cdot ^9C_3$$
or
$$\sum_{d=4}^9 {^{d-1}}C_3\cdot ^dC_3 $$
possible ways to select  $a,b,c,d,e,f,g$.
This is $7608$.
A: For $d=4$ you have to select $a,b,c$ from the set of {1,2,3}. You can do that in $C_3^3=1$ different ways. On the right side you have to select $d,e,f$ from the set of {0,1,2,3}. The number of possible choices is $C_4^3=4$. So you end up with 1x4=4 possibilities.
For $d=5$ you have to select $a,b,c$ from the set of {1,2,3,4}. You can do that in $C_4^3=4$ different ways. On the right side you have to select $d,e,f$ from the set of {0,1,2,3,4}. The number of possible choices is $C_5^3=10$. So you end up with 4x10=40 possibilities.
You can project the similar logic for $d=6,7,8,9$. So the total number is:
$$\sum_{k=4}^9\binom {k-1} 3 \binom k 3=\\\binom 3 3 \binom 4 3+\binom 4 3 \binom 5 3 +\binom 5 3 \binom 6 3+\binom 6 3 \binom 7 3+\binom 7 3 \binom 8 3+\binom 8 3 \binom 9 3=7608$$
