Wiki User
∙ 10y agoYou have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
Wiki User
∙ 10y ago9 odd numbers less than 100 can be formed. They are: 3,5,7,35,37,53,57,73 and 75.
There are 2000 such numbers.
There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.
There are 676,000 ways to make the license plates.
If repetition is allowed and leading zeros are allowed, then you have 000 to 999, which is 1,000 possibilities (I will not list them here). If no leading zeros are allowed, then you have 100 through 999, so there are 900 possibilities. If there are other restrictions, such as no repetition, then I'll refer you to the related link at MathsIsFun.com
64 if repetition is allowed.24 if repetition is not allowed.
If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
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.
-123456787
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.
10^7 if the repetition of digits is allowed. 9*8*7*6*5*4*3 , if the repetition of digits is not allowed.
125
290
-4
125 There are five choices for each of the three digits (since repetition is allowed). So there are 5*5*5=125 combinations.
Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77