It depends on whether the objects are distinguishable. I’ll assume for now that they are. First choose which $4$ of the $7$ places in the line are to be occupied by $A$ objects; this can be done in $\binom74$ ways. (By the way, if you use the $C$ notation, that’s $^7C_4$, not $^4C_7$.) Now you have $17$ ways to pick the first $A$ object, $16$ ways to pick the next $A$ object, $15$ ways to choose the third, and $14$ ways to choose the last $A$ object. After that you can choose the first $B$ object in $13$ ways, the second in $12$ ways, and the last in $11$ ways. The total number of lineups is therefore
$\binom74\cdot17\cdot16\cdot15\cdot14\cdot13\cdot12\cdot11=3,430,627,200\;.$
Added: As Henry points out in the comments, there is another straightforward way to analyze this version of the problem. There are $\binom{17}4$ ways to choose $4$ $A$ objects, $\binom{13}3$ ways to choose $3$ $B$ objects, and then $7!$ possible orders for the $7$ objects chosen, for a total of $\binom{17}4\binom{13}37!$ lineups.
If the various $A$ objects are indistinguishable from one another, as are the $B$ objects, then the only thing that distinguishes one lineup from another is the placement of the $A$ objects in the line, so there are only $\binom74=35$ lineups.
If the $A$ objects and the $B$ objects are distinguishable, so that you can tell which $4$ of the one you have and which $3$ of the other, but you don’t care about the order in which the $A$ objects occupy the $4$ $A$ slots or the order in which the $B$ objects occupy the $3$ $B$ slots, but you do care which slots in the lineup are filled with $A$ objects and which with $B$ objects, then you $\binom74$ ways to choose which $4$ slots get $A$ objects, $\binom{17}4$ ways to choose which $4$ $A$ objects to use, and $\binom{13}3$ ways to choose the $3$ $B$ objects, for a total of
$\binom74\binom{17}4\binom{13}3=23,823,800$
lineups.