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Can you list all the 6 digit combination possibilities for the numbers 1 2 3 4 8 and 9 using each number only once?
Wiki User
∙ 11y ago
Yes, I could.
-- If the sequence is significant, then there are 720 different possibilities,
and I can't list all of them here.
-- But if the sequence is not significant, then there is only one possible
6-digit group ("combination") using the digits 1, 2, 3, 4, 8, and 9.
Wiki User
∙ 11y ago
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How many numbers less than 350 can be formed from 1 2 3 4 5 6 if a digit may be repeated in a number?
138 including 1 and 2 digit numbers.-- I'll assume that we're only talking about whole numbers, and no decimals or fractions.-- The number can't have more than 3 digits. If it had four, it would be more than 1,000.-- One-digit numbers . . . 6 possibilities, all less than 350.-- Two-digit numbers . . . 6 x 6 = 36 possibilities, all less than 350.-- In three-digit numbers . . .. . . the first digit can be 1, 2, or 3.. . . If the first digit is 3, then the second one can be 1, 2, 3, or 4 = 4 choices.. . . . . . For each of the 4 choices, the third digit can be 1, 2, 3, 4, 5, or 6 = 6 choices.. . . . . . Total possibilities beginning with a 3 . . . 4 x 6 = 24. . . If the first digit is 1 or 2, then. . . . . . the second digit can be 1, 2, 3, 4, 5, or 6 = 6 choices. For each one . . .. . . . . . the third digit can be 1, 2, 3, 4, 5, or 6 = 6 choices.. . . . . . Total possibilities beginning with 1 or 2 . . . 2 x 6 x 6 = 72Adam up:One-digit possibilities . . . . . . . . . . . . . . . . . . . . . . 6Two-digit possibilities. . . . . . . . . . . . . . . . . . . . . . 36Three-digit possibilities beginning with 3. . . . . . . 24Three-digit possibilities beginning with 1 or 2 . . . 72Total. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
How many 4 digit combinations can be made from these 3 numbers 2-6-5?
Assuming you're allowed to repeat numbers, the answer is 81. For each digit, you have three choices: 2, 5, and 6. So there are three one-digit numbers you can make, and each one digit number can be combined with any of three other digits to make a two-digit number, giving 9 (3 times 3) possibilities. Now you can take any of the nine two-digit numbers as the tens and ones places in the four-digit number and any other two-digit number as the thousands and hundreds places, which gives 81 (9 times 9) possibilities. If you can't repeat numbers, the answer is zero because you don't have enough numbers to fill all the digits.
How many 3 digit numbers contain no repetition of digits?
Let's consider the following distribution: __ __ __ Each space represents one number. The first number can be anything from 0-9 (10 possibilities). The second number, after choosing the first number, can only have 9 possibilities. The reasoning behind this is if we choose the same number as the first, then there is a repetition. So we can't use that same number again. The third number can only have 8 possibilities by the same logic as for the second digit. Therefore there can be 10*9*8 = 720 3-digit numbers with no repetitions of digits.
How many six digit numbers can you find using only 1 and 0?
32...calculated as possibillities for the first digit: 1 (because if it was 0, it would only be a 5 digit number) x 2 possibilities for the second digit x 2 possibilities for the third digit...etc. ie, 1x2x2x2x2x2=32
How do i compile a list of all 5 digit numbers i can make from 7 numbers where the number can be repeated example 11233 would be a valid number?
It'll be a long list; there are 57 possibilities.
If A standard combination lock has 60 numbers on its face from 0 to 59 a combination consits of three numbers and numbers can be repeated what is the total number of possible combinations?
There are 60 possible numbers for the first number, A, in the combination (1,2,3,...,59,60).For each of outcome of A, there are 60 outcomes for the second number, B, giving a two digit combination 60x60=360 possibilities.For each of these 360 outcomes, there are 60 outcomes for the third number, C, making the number of possible combinations 360x60=21600.Or 60 possibilities x 60 possibilities x 60 possibilities = 21,600 possibilities.
How many 4 digit numbers have only odd numbers?
625. Here's how:There are five odd digits {1,3,5,7,9} so our numbers will be made using only these digits. This is the same number of possibilities as if you had a base 5 number system where the allowed digits are {0,1,2,3,4}. For a 1 digit base-5, there are 5 possibilities, for a two digit base-5 there are 5² = 25 possibilities, so for a 4-digit, there are 5^4 = 625possible. This is the same number you will have with only using odd digits.
What are all the possibilities of 4-digit numbers from numbers 0-9?
There are 10,000 possible combinations, if each number can be used more than once.
How many three digit numbers have three distinct digits?
For a 3 digit number, the left most or the most significant digit cannot be zero. So it can be 1,2,3,4,5,6,7,8 or 9 which is 9 possibilities. The middle number can be 0,1,2,3,4,5,6,7,8,9 which is 10 possibilties but one of the digits has been chosen already as the first digit, so the possibilities are only 9. The right most number can be 0,1,2,3,4,5,6,7,8,9 which is 10 possibilities but two of the digits have been already used by the left most and the middle digits. That leaves only 8 possibilities. So the total number of three digit numbers that have three distinct digits is 9 x 9 x 8 = 81 x 8 = 648 possibilities
How many numbers less than 350 can be formed from 1 2 3 4 5 6 if a digit may be repeated in a number?
138 including 1 and 2 digit numbers.-- I'll assume that we're only talking about whole numbers, and no decimals or fractions.-- The number can't have more than 3 digits. If it had four, it would be more than 1,000.-- One-digit numbers . . . 6 possibilities, all less than 350.-- Two-digit numbers . . . 6 x 6 = 36 possibilities, all less than 350.-- In three-digit numbers . . .. . . the first digit can be 1, 2, or 3.. . . If the first digit is 3, then the second one can be 1, 2, 3, or 4 = 4 choices.. . . . . . For each of the 4 choices, the third digit can be 1, 2, 3, 4, 5, or 6 = 6 choices.. . . . . . Total possibilities beginning with a 3 . . . 4 x 6 = 24. . . If the first digit is 1 or 2, then. . . . . . the second digit can be 1, 2, 3, 4, 5, or 6 = 6 choices. For each one . . .. . . . . . the third digit can be 1, 2, 3, 4, 5, or 6 = 6 choices.. . . . . . Total possibilities beginning with 1 or 2 . . . 2 x 6 x 6 = 72Adam up:One-digit possibilities . . . . . . . . . . . . . . . . . . . . . . 6Two-digit possibilities. . . . . . . . . . . . . . . . . . . . . . 36Three-digit possibilities beginning with 3. . . . . . . 24Three-digit possibilities beginning with 1 or 2 . . . 72Total. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138
How many 4 digit combinations can be made from these 3 numbers 2-6-5?
Assuming you're allowed to repeat numbers, the answer is 81. For each digit, you have three choices: 2, 5, and 6. So there are three one-digit numbers you can make, and each one digit number can be combined with any of three other digits to make a two-digit number, giving 9 (3 times 3) possibilities. Now you can take any of the nine two-digit numbers as the tens and ones places in the four-digit number and any other two-digit number as the thousands and hundreds places, which gives 81 (9 times 9) possibilities. If you can't repeat numbers, the answer is zero because you don't have enough numbers to fill all the digits.
How many 3 digit numbers contain no repetition of digits?
Let's consider the following distribution: __ __ __ Each space represents one number. The first number can be anything from 0-9 (10 possibilities). The second number, after choosing the first number, can only have 9 possibilities. The reasoning behind this is if we choose the same number as the first, then there is a repetition. So we can't use that same number again. The third number can only have 8 possibilities by the same logic as for the second digit. Therefore there can be 10*9*8 = 720 3-digit numbers with no repetitions of digits.
How many six digit numbers can you find using only 1 and 0?
32...calculated as possibillities for the first digit: 1 (because if it was 0, it would only be a 5 digit number) x 2 possibilities for the second digit x 2 possibilities for the third digit...etc. ie, 1x2x2x2x2x2=32
How do i compile a list of all 5 digit numbers i can make from 7 numbers where the number can be repeated example 11233 would be a valid number?
It'll be a long list; there are 57 possibilities.
What five digit number is greater than 20000 but less than 30000?
There are 9999 possibilities; they are the numbers from 20001 to 29999 inclusive.
How many palindromes are between 100 and 10000?
For the group of 3 digit numbers : Any number from 1-9 can be the 1st digit, any number from 0-9 can be the 2nd digit, the 3rd digit can only be one number that matches the 1st digit. The number of possibilities is thus, 9 x 10 x 1 = 90. For the group of 4 digits : Any number from 1-9 can be the 1st digit, any number from 0-9 can be the second digit, the 3rd digit can only be the one number that matches the 2nd number, and the 4th digit can only be the one number that matches the 1st digit. Number of possibilities is 9 x 10 x 1 x 1 = 90. Giving a total of 180 palindromic numbers between 100 and 10000.
How many 3-digit numbers can be formed using 0123456789 when repetition is not allowed?
-123456786
