We need the number of ordered outcomes $(x_1,x_2,x_3,x_4)$ when a die is thrown 4 times such that $x_1+x_2+x_3+x_4=16,$ where each $x_i\in\{1,2,3,4,5,6\}$.

Step 1: Convert to a nonnegative integer equation

Let $y_i=x_i-1 \quad (i=1,2,3,4).$ Then each $y_i\ge 0$ and $y_i\le 5$, and $y_1+y_2+y_3+y_4=16-4=12.$

So we need the number of nonnegative integer solutions of $y_1+y_2+y_3+y_4=12,$ with the restriction $y_i\le 5$.

Step 2: Count all nonnegative solutions without the upper bound

Ignoring the condition $y_i\le 5$, the number of nonnegative solutions is $\binom{12+4-1}{4-1}=\binom{15}{3}=455.$

Step 3: Subtract solutions where at least one variable exceeds 5

We now subtract cases where some $y_i\ge 6$.

For a particular variable, say $y_1\ge 6$, write $y_1'=y_1-6\ge 0.$ Then $y_1'+y_2+y_3+y_4=12-6=6.$ Number of solutions: $\binom{6+4-1}{3}=\binom{9}{3}=84.$

There are 4 choices for which variable is at least 6, so total to subtract initially: $4\cdot 84=336.$

Step 4: Add back intersections

If two variables are both at least 6, say $y_1,y_2\ge 6$, then let $y_1'=y_1-6,\quad y_2'=y_2-6.$ This gives $y_1'+y_2'+y_3+y_4=12-12=0.$ So there is exactly $\binom{3}{3}=1$ solution for each such pair.

Number of pairs: $\binom{4}{2}=6.$ So we add back $6\cdot 1=6.$

Step 5: Higher intersections

Three variables cannot each be at least 6, because that would require sum at least $18>12$. So no further terms occur.

Step 6: Final count

By inclusion-exclusion, $455-336+6=125.$

Therefore, the number of ways is $\boxed{125}.$