3. If three numbers are taken simultaneously from the 14 integers $1,2, \cdots, 14$, such that the absolute difference between any two numbers is not less than 3, then the number of different ways to choose is $\qquad$
3. 120 .

Let the three integers taken out be $x, y, z (x<y<z)$.
$$
\begin{array}{l}
\text { Let } a=x, b=y-x-2, \\
c=z-y-2, d=15-z .
\end{array}
$$

Thus, $a, b, c, d \geqslant 1$.
If $a, b, c, d$ are determined, then $x, y, z$ are uniquely determined.
Since $a+b+c+d=11$, it is equivalent to dividing 11 identical balls into four piles.
Using the stars and bars method, we get $C_{10}^{3}=120$.