Question

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) 720

I got this question in an interview for job.

My query is from Permutations-2 in portion Permutations and Combinations of Mathematics – Class 11

Answer 1

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.