Answer to Question #132814 in Combinatorics | Number Theory for AY YAWA

Question

Given digits 0,1,2,3,4,5,6 & 7

How many 4'digit different codes are possible if repetition is not allowed

Answer 1

There are 8possibilities from 0 to 7. but for the first digit code there are 8possiblities, the second will have 7possiblites, with the third having 6possibilities and the fourth will have 5possibilities so

=8 x 7 x 6 x 5 = 1680

A total of 1680 of 4digit codes can be allowed