^nP0 = ________________(a) n!(b) 1(c) \(\frac{1}{(n)!}\)(d) (n-1)!I got this question in an online interview.My doubt is from Permutations-1 topic in division Permutations and Combinations of Mathematics – Class 11

^nP0 = ________________(a) n!(b) 1(c) \(\frac{1}{(n)!}\)(d) (n-1)!I go

1.	^nP0 = ________________(a) n!(b) 1(c) \(\frac{1}{(n)!}\)(d) (n-1)!I got this question in an online interview.My doubt is from Permutations-1 topic in division Permutations and Combinations of Mathematics – Class 11
Answer» The CORRECT answer is (B) 1 The explanation: We know, ^nPr = \(\FRAC{N!}{(n-r)!}\). ^nP0 = \(\frac{n!}{(n-0)!} = \frac{n!}{(n)!}\) = 1.

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