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Q: How many different two digit numbers can you form using the digits 1 2 5 7 8 and 9 without repetition?

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

15 of them.

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

24

Six (6)

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.

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.

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.

9*8*7 = 504 of them.

3*2*1 = 6 of them.

Assuming that the first digit can't be zero: If repetition of digits is permitted: (5 x 6 x 6) = 180 numbers of 3 digits. If repetition of digits is not permitted: (5 x 5 x 4) = 100 numbers of 3 digits.

If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.

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