0
If using numbers 1 through 9 how many different 4 digit combinations can be made with no restrictions on order or repeats?
Wiki User
∙ 7y ago
Each digit can appear in each of the 4 positions.
There are 9 digits, therefore there are 9⁴ = 6561 such combinations.
Wiki User
∙ 7y ago
Jack Kolars
Add your answer:
Earn +20 pts
How many 3 digit combinations can be formed using 0 5 7 8?
Assuming you are using combinations in the colloquial way (which is the mathematical "permutations" where order of selection does matter) to create a 3 digit number that does not start with 0, ie creating a number that is between 100 and 999 inclusive then: If repeats are not allowed there are 3 × 3 × 2 = 18 possible numbers If repeats are allowed, then there are 3 × 4 × 4 = 48 possible numbers. If you are using combinations in the mathematical sense where order of selection does not matter and are creating groups of 3 digits, then: If repeats are not allowed there are 4 possible groups If repeats are allowed there are 20 possible groups.
How do you generate a list of five digit combinations of only the numbers 1 2 3 4 and 5 with out repeats?
you could make a probability tree if you could be bothered
How many 6 number combinations can you make with numbers 1-50. No 1 number repeats?
The number of combinations is 50C6 = 50*49*48*47*46*45/(6*5*4*3*2*1) = 15,890,700
How many different 2 digit numbers can be formed from the digits 2 6 7 and 8?
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
How many 8 number combinations are their using numbers 0 -9?
there are 10 numbers that you can choose from 0-9 you are choosing 8 at a time, meaning no repeats so the answer is "10 choose 8" or 10!/(8!*(8-2)!)
If using numbers 0 through 9 how many different 4 digit combinations can be made with no restrictions on order or repeats?
In other words, how many 4 digit combination locks are there using the digits 0-9 on each wheel. There are 10×10×10×10 = 10⁴ = 10,000 such combinations.
How many different ways can I place 4 numbers?
Without repeats, 24. With repeats, 256.
How many 3 digit combinations can be formed using 0 5 7 8?
Assuming you are using combinations in the colloquial way (which is the mathematical "permutations" where order of selection does matter) to create a 3 digit number that does not start with 0, ie creating a number that is between 100 and 999 inclusive then: If repeats are not allowed there are 3 × 3 × 2 = 18 possible numbers If repeats are allowed, then there are 3 × 4 × 4 = 48 possible numbers. If you are using combinations in the mathematical sense where order of selection does not matter and are creating groups of 3 digits, then: If repeats are not allowed there are 4 possible groups If repeats are allowed there are 20 possible groups.
How many combinations can be made from 1234567890?
If repeats are allowed than an infinite number of combinations is possible.
How is a rational number that repeats different from a irratrional number that repeats?
Irrational numbers can not be repeating decimals. Any number that is a repeating decimal is rational.
How do you generate a list of five digit combinations of only the numbers 1 2 3 4 and 5 with out repeats?
you could make a probability tree if you could be bothered
How many 6 number combinations can you make with numbers 1-50. No 1 number repeats?
The number of combinations is 50C6 = 50*49*48*47*46*45/(6*5*4*3*2*1) = 15,890,700
How many different 2 digit numbers can be formed from the digits 2 6 7 and 8?
Without repeats there are 4 × 3 = 12 possible 2 digit numbers. With repeats there are 4 × 4 = 16 possible 2 digit numbers.
How many 8 number combinations are their using numbers 0 -9?
there are 10 numbers that you can choose from 0-9 you are choosing 8 at a time, meaning no repeats so the answer is "10 choose 8" or 10!/(8!*(8-2)!)
How many 6 number combinations can made from 50 numbers 1-50?
We are not told anything about repeats so I will assume you can repeat a digit. For example, you can have 111111 The first digits has 50 choices so does the second and the third..etc so we have 50^6 combinations.
How many 3 digit even numbers can be formed by using numbers 01234?
Assuming leading zeros are not permitted, then: If repeats are not allowed there are 30 possible numbers. If repeats are allowed there are 60 possible numbers.
How many odd 3- digit positive integers can be written using the digits 1 2 4 7 and 8?
If repeats are permitted: 2 x 5 x 5 = 50 different odd numbers If repeats are not permitted: 2 x 4 x 3 = 24 different odd numbers
