Wiki User
∙ 2010-10-04 14:28:011000
Wiki User
∙ 2010-10-04 14:28:015040
-4
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.
125 There are five choices for each of the three digits (since repetition is allowed). So there are 5*5*5=125 combinations.
Just six numbers... 345, 354, 435, 453, 534 & 543
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
5040
5 ^12
290
There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.
-4
If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.
64 if repetition is allowed.24 if repetition is not allowed.
Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77
64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.64 if repetition is allowed, 6 otherwise.
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.
if the repetition is not allowed then the arrangement can be 3*2*1 = 6 ways if repetition is allowed the ways can be 6*6*6= 216 ways