Reqd no. is 9 -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 7 digits..
Ans: (5C1)^7 * 5
ie 5^8 = 390625
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
There are 2000 such numbers.
64 if repetition is allowed.24 if repetition is not allowed.
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.
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
There are 2000 such numbers.
64 if repetition is allowed.24 if repetition is not allowed.
There are 2000 such numbers.
5 ^12
If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.
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