Number in this range will be 3-digit number.

$$N=\overline{a b c}$$ such that $$a+b+c=14$$

Also, $$a \geq 1, \quad a, b, c \in\{0,1,2, \ldots 9\}$$

Case I

All 3-digit same

$$\Rightarrow 3 a=14$$ not possible

Case II

Exactly 2 digit same:

$$\Rightarrow 2 a+c=14$$

$$\begin{aligned} & (a, c) \in\{(3,8),(4,6),(5,4),(6,2),(7,0)\} \\ & \Rightarrow\left(\frac{3!}{2!}\right) \text { ways } \Rightarrow 5 \times 3-1 \\ & =15-1=14 \end{aligned}$$

Case III

All digits are distinct

$$a+b+c=14$$

without losing generality $$a > b > c$$

$$\begin{aligned} & (a, b, c) \in\left\{\begin{array}{l} (9,5,0),(9,4,1),(9,3,2) \\ (8,6,0),(8,5,1),(8,4,2) \\ (7,6,1),(7,5,2),(7,4,3) \\ (6,5,3) \end{array}\right. \\ & \Rightarrow 8 \times 3!+2(3!-2!)=48+8=56 \\ & =0+14+56=70 \end{aligned}$$