Q: Using each digit only once list all possible 3 digit numbers that can be made with 2 or 4 or 7?

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-123456786

This is not possible, since there are only five single digit odd numbers, which are 1, 3, 5, 7 and 9.

24 different numbers.

It is .56789

998,000

Related questions

-123456786

This is not possible, since there are only five single digit odd numbers, which are 1, 3, 5, 7 and 9.

There are 10000 such codes. Each of the numbers 0-9 can be in the first position. With each such first digit, each of the numbers 0-9 can be in the second position. With each such pair of the first two digits, each of the numbers 0-9 can be in the third position. etc.

Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.

There are 720 of them. The three digit counting numbers are 100-999. All multiples of 5 have their last digit as 0 or 5. There are 9 possible numbers {1-9} for the first digit, There are 10 possible numbers {0-9} for each of the first digits, There are 8 possible numbers {1-4, 6-9} for each of the first two digits, Making 9 x 10 x 8 = 720 possible 3 digit counting numbers not multiples of 5.

ten factorial = 10! = 3,628,800

six

24 different numbers.

Any 4 from 10 in any order = 10 x 9 x 8 x 7 = 5040

It is .56789

If you want 4-digit numbers, there are 24 of them.

Proceed from left to right, using the largest possible digit in each position.