Each of the five persons $A,B,C,D\text{ and }E$ possesses an unequal number (less than $10$) of similar items. $A,B\text{ and }C$ possess $21$ items in all, while $C,D\text{ and }E$ possess $7$ items in all. How many items do $A$ and $B$ possess in all?

Given that each person has a distinct number of items which is less than $10$. So, the allowable individual item counts are non-negative distinct integers from the set $\{0,1,2,3,4,5,6,7,8,9\}$.
We have:

1. $A+B+C=21$
2. $C+D+E=7$

Let us analyze equation (2). Since $C,D,E$ are distinct non-negative integers, the minimum possible sum of two distinct numbers from $D$ and $E$ when $C$ varies can be considered.
The three numbers $C,D,E$ must be distinct integers from $\{0,1,2,3,4,5,6,7,8,9\}$ whose sum is $7$. Let's look at the maximum value $C$ can take.
If $C=5$, $D+E=2⟹\{0,2\}$ (since $D\ne E$).
If $C=4$, $D+E=3⟹\{0,3\}$ or $\{1,2\}$.
If $C=3$, $D+E=4⟹\{0,4\}$ (since $\{1,3\}$ repeats $3$).

Now let's check equation (1): $A+B=21-C$.
Since $A$ and $B$ are distinct and both are $<10$, the maximum value that $A+B$ can possibly reach is $9+8=17$.
Therefore:

$$21-C\le 17⟹C\ge 4$$

From our options for equation (2), $C$ can be $4$ or $5$.
Case 1: If $C=5$, then $A+B=21-5=16$. Let's check if $A$ and $B$ can be distinct numbers $<10$: $9+7=16$. This gives the set of numbers as $\{9,7,5\}$ for $A,B,C$. For $D,E$, since $C=5$ and $D+E=2$, the numbers are $\{0,2\}$. The total set of five values is $\{9,7,5,0,2\}$, which are all distinct and $<10$.
Case 2: If $C=4$, then $A+B=21-4=17$. The only way to get $17$ with distinct integers less than $10$ is $9+8=17$. This gives the values $\{9,8,4\}$ for $A,B,C$. For $D,E$, since $C=4$ and $D+E=3$, the choices are $\{0,3\}$ or $\{1,2\}$. Both configurations give all $5$ elements distinct and $<10$.

Let's check the options listed for $A+B$: $15,17,18$. The value $17$ matches Case 2 perfectly.
Thus, $A$ and $B$ possess $17$ items in all.