An engineer is required to visit a factory for exactly four days during the first 15 days of every month and it is mandatory that no two visits take place on consecutive days. Then the number of all possible ways in which such visits to the factory can be made by the engineer during $1-15$ June 2021 is ______.

$\text{Given: exactly } 4 \text{ days chosen from } 15$

$\text{no two chosen days are consecutive}$

$\text{To find: number of ways to pick } 4 \text{ non-consecutive days from } 1 \text{ to } 15$

$\text{Plan: } 1.\ \text{model the } 4 \text{ chosen days}$

$2.\ \text{shift to remove the "no consecutive" condition}$

$3.\ \text{count with a binomial coefficient}$

$1 \le d_1 < d_2 < d_3 < d_4 \le 15$

$d_{i+1} \ge d_i + 2 \quad (i = 1,2,3)$

$e_i = d_i - (i-1)$

$e_{i+1} - e_i = (d_{i+1} - d_i) - 1 \ge 2 - 1 = 1$

$e_1 \ge 1,\qquad e_4 = d_4 - 3 \le 15 - 3 = 12$

$\{\text{non-consecutive } 4\text{-subsets of } 1..15\} \longleftrightarrow \{4\text{-subsets of } 1..12\}$

$\text{Number of ways} = \binom{12}{4}$

$\binom{12}{4} = \dfrac{12 \cdot 11 \cdot 10 \cdot 9}{4 \cdot 3 \cdot 2 \cdot 1}$

$= 11 \cdot 5 \cdot 9 = 495$

$d = (1,3,5,15) \Rightarrow e = (1,2,3,12)$

$\binom{12}{4} = 495$