Permutations & Combinations
Permutation and Combination
star_batch_jee_advanced_2025
Grade None
$P = n(n^2-1)(n^2-4)(n^2-9)\cdots(n^2-100)$ is always divisible by: $(n \in I)$
Step-by-Step Solution
Key Concept: $P$ represents a product of 21 consecutive integers centered at $n$, and any such product is divisible by $21!$ and all its factors.
We have $P = n(n^2-1)(n^2-4)(n^2-9)\cdots(n^2-100) = n(n-1)(n+1)(n-2)(n+2)(n-3)(n+3)\cdots(n-10)(n+10)$, which is a product of 21 consecutive integers from $(n-10)$ to $(n+10)$. Any product of 21 consecutive integers is divisible by $21! = 21 \times 20 \times \cdots \times 1$. Since $21!$ contains all factors of $2!, 3!, 4!, 5!, 6!$, $(5!)^4$, and $(10!)^2$ as divisors, $P$ is divisible by all these. For option 4, $10! \mid 21!$ and $11! \mid 21!$, so both divide $P$.
Correct Answer: 1,2,3,4