What is the probability that the maximum number of shots fired successively from a type A gun is $2$? A gun salute always takes place at the funeral of a military leader who has died in a certain country. 
(The $21$ gun salute where $21$ rounds are fired - is the most common for the most senior military leaders and the number of rounds progressively gets smaller as the rank gets lower.) 
Lance, who was an air marshal has died and his funeral is there tomorrow. At his funeral, a $14$ gun salute will be held (a total of $14$ rounds will be fired in succession) comprising of $5$ rounds from a Type A gun, $3$ rounds from a Type B gun and the rest from a Type C gun. => $6$ right?
If the guns are fired randomly, what is the probability that the maximum number of shots fired successively from a type A gun is $2$?
Answer: $\frac{121968}{168168} = \frac{66}{91} =0.7253$
I was trying to take the approach of taking the total number of possible ways ($\frac{14!}{{5!}{3!}{6!}} = 168168$) and subtracting the ways where you take all the type A out and then place them back in but in between others in ways that there are more than 2 successive fires.
I got confused because I don't know how to account for the repeated type A after putting them back in in such a fashion.
I had:
$3$ A's in succession and $2$ A's in succession
$3$ A's in succession and $2$ A's not in succession
$4$ A's in succession and $1$ A
$5$ A's in succession
I tried to approach it like this:
I'm just going to show you what I did to attempt to get the numerator of the answer ($121968$) before dividing it by $168168$ (obviously to get the probability).
$168168 - \frac{{9!}{10\choose2}{3!}{2!}{2!}}{5!3!6!} - \frac{{9!}{10\choose3}{3!}{3!}}{5!3!6!} - \frac{{9!}{10\choose2}{4!}{2!}}{5!3!6!} - \frac{{9!}{10\choose1}{5!}}{5!3!6!}$
$= 168168 - 756 - 3024 - 1512 - 840 = 162036$
What am I doing wrong? How do they get $121968$

Your expression is mostly right. Take the case "3 A's in succession and 2 A's in succession". We have 2 subcases, either the AAA comes first or the AA comes first. These are equivalent.
Let X denote a B or a C (whether it's B or C, we decide at the end). We have 9 X's to place in 3 slots as shown:
 AAA AA
^   ^  ^

Except we must place an X in the middle slot. So we have 8 X's and 2 blocks of A's to arrange; one possible arrangement is (X AAA(X) X X X AA X X X X). Total # of ways to do this is 10C2.
Finally we choose B's and C's among the X's. There are 9C3 ways to do this. So the total # of ways for this case is $2\binom{10}{2}\binom{9}{3}$. The other cases are similar.
