Edit: My first answer (appended in brackets) is much too complicated. Here is the final version:
The given hexagonal array $H$ of beads can be viewed (in two ways) as the union of three parallelogram shaped arrays containing $ab$, $bc$, and $ca$ beads respectively, any two of them sharing an edge of length $a$, $b$, or $c$, and all three of them having one bead in the interior of $H$ in common. By the inclusion-exclusion principle the total number $N$ of beads is therefore given by $$N=(ab + bc + ca) -(a+b+c)+1\ .$$
[Assume $a\leq b\leq c$, where $a$ is the number of tiles on the vertical edge. Partition the figure by two auxiliary vertical lines into two trapezoidal parts and a central parallelogram shaped part. The $b$ vertical rows in the trapezoidal parts contain resp. $a$, $a+1$, $\ldots$, $a+b-1$ hexagons, and the $c-b-1$ "inner" vertical rows of the parallelogram shaped part contain $a+b-1$ hexagons each. (When $c=b$ then $c-b-1=-1$ which takes care of the fact that in this case the longest vertical column, of length $a+b-1$, has been counted twice so far.) Therefore the total number $N(a,b,c)$ of tiles is given by $$\eqalign{N(a,b,c)&=\bigl(a+(a+1)+\ldots+(a+b-1)\bigr)+(c-b-1)(a+b-1)\cr &= b\bigl(a+(a+b-1)\bigr)+(c-b-1)(a+b-1)\cr &=(ab+bc+ca)-(a+b+c)+1\ .\cr}$$ Since the resulting total is a symmetric function of $a$, $b$, $c$ the assumption $a\leq b\leq c$ used in the proof turns out to be unnecessary.]