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There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.

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Q: How many different three digit numbers can you form using the digits 1 2 3 5 6 7 and 9 without repetition?
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Related questions

How many different two digit numbers can you form using the digits 1 2 5 7 8 and 9 with repetition?

30 without repetition (6P2) 66 with repetition (12C2)


How many numbers can be formed using the digits 37 and 9 without repetition?

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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 four digits numbers can be made with 1 4 5 9 without repetition?

24


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How many different numbers can you form using all four numbers of 1 9 8 4?

The four digits can be used to produce infinitely many different numbers if repetition is permitted. Without repetition, there are 24 possible numbers. A lot more can be produced if the numbers are combined using binary oprations, fore example, 19 * 8/4 = 19*2 = 38.


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.


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You 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.


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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?

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