In base 11, the digits range from 0 to 10, and in base 12, the digits range from 0 to 11.
So, we can rewrite $249_{11}$ as $2\cdot11^2+4\cdot11^1+9\cdot11^0$ and $3AB_{12}$ as $3\cdot12^2+10\cdot12^1+11\cdot12^0$.
Evaluating these expressions, we get $2\cdot121+4\cdot11+9\cdot1=242+44+9=295$
and $3\cdot144+10\cdot12+11\cdot1=432+120+11=563$.
So, $249_{11}+3AB_{12} = 295 + 563 = \boxed{858}$.
The answer is: 858
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Express $249_{11}+3AB_{12}$ as a base 10 integer. Here $A$ and $B$ denote the digits whose values are 10 and 11, respectively, in base 12.

### Answer:
