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The factorial of a non-negative integer $n$, denoted $n!$, is defined as
A$n \cdot (n+1) \cdot (n+2) \cdots$
B$n \cdot n$
C$n + (n-1) + (n-2) + \ldots + 1$
D$1 \cdot 2 \cdot 3 \cdots n$, with $0! = 1$
Answer & Solution
Correct answer: D. $1 \cdot 2 \cdot 3 \cdots n$, with $0! = 1$
1. NCERT §6.2 (Factorial Notation): $n!$ is the PRODUCT of all positive integers from 1 to $n$ inclusive.
2. $n! = 1 \cdot 2 \cdot 3 \cdots n$.
3. By CONVENTION: $0! = 1$ (this makes formulas with $n!$ work consistently when $n = 0$).
4. Examples: $5! = 120$, $10! = 3{,}628{,}800$.
5. RECURSIVE definition: $n! = n \cdot (n-1)!$, which gives the same values.
6. Option A goes upward (infinite product). Option C ignores the product over all integers. Option D is the SUM, not product.
_Source: NCERT Class 11 Mathematics, Ch 6 "Permutations and Combinations", §6.2 (Factorial Notation), p. 2–3._
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