We need to choose exactly $4$ visit days from the first $15$ days, with the condition that no two chosen days are consecutive.

So the problem is:

Choose $4$ numbers from $\{1,2,3,\dots,15\}$ such that no two are consecutive.

Use the standard transformation for selections with no consecutive integers.

If the chosen days are $a_1<a_2<a_3<a_4,$ with $a_{i+1}-a_i\ge 2,$ then define $b_1=a_1,\quad b_2=a_2-1,\quad b_3=a_3-2,\quad b_4=a_4-3.$

Then:

• $b_1<b_2<b_3<b_4$
• each $b_i$ is an integer
• and the largest possible value of $b_4$ is $15-3=12$

So now we just need to choose $4$ distinct numbers from $\{1,2,3,\dots,12\}$.

Therefore, the number of ways is $\binom{12}{4}.$

Compute it: $\binom{12}{4}=\frac{12\cdot 11\cdot 10\cdot 9}{4\cdot 3\cdot 2\cdot 1}=495.$

Hence, the total number of possible schedules is $495.$

Comparison with stored answer:

• Derived answer: $495$
• Stored correct answer: $495$
• They match.