Vern has a collection of pennies, nickels, and dimes. The ratio of the number of pennies to the number of nickels is $9:2,$ and the ratio of the number of nickels to the number of dimes is $3:4.$ If the total worth of Vern's collection is $\$10.96,$ then how many coins do they have in total?

Let $p$ be the number of pennies, $n$ be the number of nickels, and $d$ be the number of dimes that Vern has. The given ratios tell us that $n=\dfrac{2}{9}p$ and $d=\dfrac{4}{3}n.$ Therefore, $d=\dfrac{4}{3} \cdot \dfrac{2}{9}p=\dfrac{8}{27}p,$ so we have

\[p:n:d=1:\dfrac{2}{9}:\dfrac{8}{27}.\]To turn this into a ratio of integers, we multiply every part of the ratio by $27.$ Doing this, we see that $p:n:d=27:6:8.$

Therefore, we can think of Vern's collection as consisting of several groups, each of which contains $27$ pennies, $6$ nickels, and $8$ dimes. Let $x$ be the number of such groups of coins that Vern has. Then Vern has $27n$ pennies, $6n$ nickels, and $8n$ dimes. Since a penny is worth $1$ cent, a nickel is worth $5$ cents, and a dime is worth $10$ cents, the total worth of Vern's coins in cents is $27n \cdot 1 + 6n \cdot 5 + 8n \cdot 10.$ However, we know Vern has $\$10.96,$ or $1096,$ cents, so we can write an equation:

\[27n \cdot 1 + 6n \cdot 5 + 8n \cdot 10=1096.\]Simplifying the left-hand side, we get $137n=1096.$ Dividing both sides by $137,$ we get $n=8.$ This tells us that Vern has $8$ groups of $27+6+8=41$ coins, for a total of $8 \cdot 41=\boxed{328}$ coins.