Q: How many different three-digit whole numbers are possible using the digits 1 and 2 and 3 and 4 and 5?

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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.

There are 2 possible digits for the first digit (3 or 4), leaving 3 possible digits for the second digit (5 and 6 and whichever was not chosen for the first), leaving 2 possible digits for the third. Thus there are 2 × 3 × 2 = 12 possible 3 digit numbers.

Interesting.Social Security numbers all have the form: (3 digits) - (2 digits) - (4 digits).That's 9 digits altogether. If you ignore the dashes, you get: xxx,xxx,xxx .With 9 places, there are 1 billion possible different numbers.

The sum is 22 times the sum of the three digits.

89,999 different numbers i guess

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Assuming you mean permutations of three digits, then the set of numbers that can be made with these digits is: 345 354 435 453 534 543 There are six possible permutations of three numbers.

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.

There are 2 possible digits for the first digit (3 or 4), leaving 3 possible digits for the second digit (5 and 6 and whichever was not chosen for the first), leaving 2 possible digits for the third. Thus there are 2 × 3 × 2 = 12 possible 3 digit numbers.

Interesting.Social Security numbers all have the form: (3 digits) - (2 digits) - (4 digits).That's 9 digits altogether. If you ignore the dashes, you get: xxx,xxx,xxx .With 9 places, there are 1 billion possible different numbers.

There are 9 possible numbers for the first digit (one of {1, 2, ..., 9}); with 9 possible digits for the second digit (one of {0, 1, 2, ..., 9} which is not the first digit)); with 8 possible digits for the third digit (one of {0, 1, 2, ..., 9} less the 2 digits already chosen); This there are 9 × 9 × 8 = 648 such numbers.

The sum is 22 times the sum of the three digits.

With the constraints you have listed, there are only 6 possible phone numbers:202-1178202-1187202-1718202-1781202-1817202-1871The first four digits are fixed, and there are six possible permutations of three different digits.

89,999 different numbers i guess

It is possible to create infinitely many numbers, of infinitely many different lengths, using the digits of the given number. Using each of the digits, and only once, there are 5! = 120 different permutations.

There are 7,290 different 4-digit numbers that can be formed from the digits 1-9 without repetition.

6 possible 3 digit combonations

Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.