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The number of PERMUTATIONS of $n$ distinct objects taken $r$ at a time is denoted $P(n,r)$ or $^nP_r$ and equals
A$\dfrac{n!}{r!}$
B$\dfrac{n!}{(n-r)!}$
C$\dfrac{n!}{r!(n-r)!}$
D$n \cdot r$
Answer & Solution
Correct answer: B. $\dfrac{n!}{(n-r)!}$
1. NCERT §6.3 (Permutations): a permutation is an ORDERED arrangement.
2. For the first position: $n$ choices. Second: $n-1$. Third: $n-2$. $\ldots$ Position $r$: $n - r + 1$ choices.
3. Total: $n \cdot (n-1) \cdot (n-2) \cdots (n-r+1) = \dfrac{n!}{(n-r)!}$.
4. Example: $P(5, 3) = 5!/(5-3)! = 120/2 = 60$.
5. Option A divides by $r!$ — that's the COMBINATION formula. Option C is the combination. Option D is wrong dimensions.
_Source: NCERT Class 11 Mathematics, Ch 6, §6.3 (Permutations — formula derivation), p. 5–7._
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