Permutations & Combinations
Permutation and Combination
star_batch_jee_advanced_2025
Grade 11
Question:
If $n$ objects are arranged in a row, then the number of ways of selecting three of these objects so that no two of them are next to each other is:
\frac{(n-2)(n-3)(n-4)}{6}
^{n-2}C_3
^{n-3}C_3 + ^{n-3}C_2
None of these
Step-by-Step Solution
Key Concept: When selecting non-adjacent objects, place them in gaps created by remaining objects, reducing the problem to choosing positions in $n-2$ available slots.
To select 3 objects from $n$ objects such that no two are adjacent, we can use the stars and bars method. Remove the 3 selected objects first, leaving $n-3$ objects. We need to place 3 objects in the $n-2$ gaps (including ends) created by these $n-3$ objects such that each gap has at most one object. This is equivalent to choosing 3 gaps from $n-2$ available gaps, giving $\binom{n-2}{3} = \frac{(n-2)(n-3)(n-4)}{6}$. Alternatively, using the complementary approach with $\binom{n-3}{3} + \binom{n-3}{2}$ also yields the same result through the hockey stick identity.
Correct Answer: 1,2,3