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This permutation problem depends on whether the numbers are allowed to be repeated. If they are, there are a possible 9999 numbers, starting with 0001 and running sequentially to 9999. If they are not allowed to be repeated, there are a possible 5040 combinations.

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Q: Which 4-digit numbers can be formed by the digits 1 2 3 4 5 6 7 8 9 0?
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Related questions

How many three digit numbers can be formed using the digits 0 to 9?

There are 900 of them.


How many different four-digit numbers can be formed from the digits 0 through 9?

10,000


How many different 5 digit numbers can be formed from the digits 0 to 9?

100000 - including numbers with leading 0s


How many 5 digits can be formed using the digits 0 2 3 4 5 when repetition is allowed that the number formed is divisible by 2 and 5 or both?

There are 2000 such numbers.


How many odd numbers less than 100 can be formed by using the digits 0 3 5 7 when repetition fo digits is not allowed?

9 odd numbers less than 100 can be formed. They are: 3,5,7,35,37,53,57,73 and 75.


How many 3 digit numbers divisible by 9 can be formed from the digits 0123456789?

There are 76 such numbers. Eight more if you allow numbers to start with 0.


How many 6-digit numbers can be formed using the digits 0 1 2 3 4 5 6 7 8 9 if digits can be repeated?

151200 numbers.


How many different 4-digit numbers can be formed from the digits of 0 2 4 and 7?

192, including ones containing repeat digits.


What odd numbers that can be formed using each of the digits 0 1 and 3 only once?

103 and 301 .


What Even numbers that can be formed using each of the digits 0 1 and 3 only once?

"130" and "310" .


How many 3 digit numbers can be formed from the digits 0 through 9 in a set of three?

If the digits can be used more than once, then 900. If not, then 648.


How many 4 digit numbers can be formed using the digits 0 1 2 3 4 5 6 if repetitions of digits are allowed?

2,401