The number of ways of getting a sum 16 on throwing a dice four times is ______

We need to find the number of ways of getting a sum of 16 when throwing a die four times.

We need the number of solutions to $$x_1 + x_2 + x_3 + x_4 = 16$$ where each $$x_i \in \{1, 2, 3, 4, 5, 6\}$$.

Let $$y_i = x_i - 1$$, so $$0 \leq y_i \leq 5$$. The equation becomes:

$$ y_1 + y_2 + y_3 + y_4 = 16 - 4 = 12 $$

Without upper bound constraint: The number of non-negative integer solutions to $$y_1 + y_2 + y_3 + y_4 = 12$$ is:

$$ \binom{12 + 3}{3} = \binom{15}{3} = \frac{15 \times 14 \times 13}{6} = 455 $$

Subtract: at least one $$y_i \geq 6$$. If $$y_1 \geq 6$$, let $$z_1 = y_1 - 6 \geq 0$$. Then $$z_1 + y_2 + y_3 + y_4 = 6$$, giving $$\binom{9}{3} = 84$$ solutions. By symmetry, there are 4 such cases:

$$ \binom{4}{1} \times \binom{9}{3} = 4 \times 84 = 336 $$

Add back: at least two $$y_i \geq 6$$. If $$y_1 \geq 6$$ and $$y_2 \geq 6$$, let $$z_1 = y_1 - 6, z_2 = y_2 - 6$$. Then $$z_1 + z_2 + y_3 + y_4 = 0$$, giving $$\binom{3}{3} = 1$$ solution. There are $$\binom{4}{2} = 6$$ such pairs:

$$ \binom{4}{2} \times \binom{3}{3} = 6 \times 1 = 6 $$

Three or more $$y_i \geq 6$$: This would require the sum to be at least 18, which exceeds 12. So this is impossible.

$$ N = 455 - 336 + 6 = 125 $$

The answer is $$\boxed{125}$$.