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Let's consider the following distribution:
__ __ __
Each space represents one number.
The first number can be anything from 0-9 (10 possibilities).
The second number, after choosing the first number, can only have 9 possibilities. The reasoning behind this is if we choose the same number as the first, then there is a repetition. So we can't use that same number again.
The third number can only have 8 possibilities by the same logic as for the second digit.
Therefore there can be 10*9*8 = 720 3-digit numbers with no repetitions of digits.
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How many 13 digit numbers are possible by using the digits 12345 which are divisible by 4 if repetition of digits is not allowed?
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
How many three digit numbers can be formed using the digits 873 and 6 without repetition?
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
How many 3 digit numbers can be formed from the digits 1234?
64 if repetition is allowed.24 if repetition is not allowed.
Where d represents a digit from 0 9 how many different ISBN numbers are possible if repetition of digits is 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.
How many 3-digit numbers can be formed by the digits 4 5 9 without repetition?
Six (6)
