JEE Advanced 2020 — Paper 2 — Question 46

We need to select 4 non-consecutive days from 15 days. Let the selected days be $d_1 < d_2 < d_3 < d_4$. Define new variables: $x_1 = d_1$, $x_2 = d_2 - d_1 - 1$, $x_3 = d_3 - d_2 - 1$, $x_4 = d_4 - d_3 - 1$, $x_5 = 15 - d_4$. The non-consecutive condition forces $x_2, x_3, x_4 \ge 1$. Let $y_2 = x_2 - 1$, $y_3 = x_3 - 1$, $y_4 = x_4 - 1$, with $y_i \ge 0$. Then $x_1 + y_2 + y_3 + y_4 + x_5 = 15 - 3 = 12$, where all variables are non-negative integers. Number of solutions = $\binom{12+5-1}{5-1} = \binom{16}{4} = 1820$. Wait, the above is incorrect. Correct approach: Use the gap method. Place 11 indistinguishable gaps (days not visited) in a row, creating 12 gaps (including ends) to insert 4 visits such that no two are consecutive. Number of ways = $\binom{12}{4} = 495$. Thus the number of possible schedules is 495.