Wiki User
∙ 2011-01-17 15:22:00Reqd no. is 13-digits.. available digits are 1 2 3 4 5.. Numbers which are divisible by 4 can be determined by the last two digits.. from the given combinations,, the numbers div by 4 are 12, 24, 32, 44, 52
so.. the number's last two digits can be any of the above 5.
Hence, we have to calculate the combinations for the first 11 digits..
Ans: (5C1)^11 * 5
i.e 5^12
Wiki User
∙ 2011-01-17 15:22:00If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
Assuming 9 numbers chosen from 56, with no repetition allowed, there are 7575968400 possible combinations.
If repetition is allowed . . . . . 343 If repetition is not allowed . . . . . 210
64 if repetition is allowed.24 if repetition is not allowed.
There are 504 of them.
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
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.
290
if the repetition is allowed the there is 6*6*6 possible ways = 216
Assuming 9 numbers chosen from 56, with no repetition allowed, there are 7575968400 possible combinations.
If repetition is allowed . . . . . 343 If repetition is not allowed . . . . . 210
-4
64 if repetition is allowed.24 if repetition is not allowed.
There are 2000 such numbers.
When the repetition is allowed: 4*4*4 = 64 numbers. Without repetition: 4*3*2 = 24 numbers.
There are 2000 such numbers.
There are 504 of them.