How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9?(a) 504(b) 729(c) 1000(d) 720I got this question in an interview for job.My query is from Permutations-2 in portion Permutations and Combinations of Mathematics – Class 11

How many 3-digit numbers are possible using permutations without repet

1.	How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9?(a) 504(b) 729(c) 1000(d) 720I got this question in an interview for job.My query is from Permutations-2 in portion Permutations and Combinations of Mathematics – Class 11
Answer» Right choice is (a) 504 Easiest explanation: 1-9 digits which MEANS 9 digits are possible. We have to ARRANGE 3 digits at a time out of 9 digits WITHOUT repetition. So, TOTAL permutations are ^nPr = ^9P3 = \(\frac{9!}{(9-3)!} = \frac{9!}{6!}\) = 9.8.7 = 504.

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How many 3-digit numbers are possible using permutations without repetition of digits if digits are 1-9?(a) 504(b) 729(c) 1000(d) 720I got this question in an interview for job.My query is from Permutations-2 in portion Permutations and Combinations of Mathematics – Class 11

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