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How many different 4 digit combinations can be made using the digits 0 to 9?
Well you have 10 possible numbers for the fist column(0-9) and 10 for the second, third, and fourth then you multiply those numbers. 10*10*10*10=10000 or 10^4=10,000. So there are 10,000 different combinations.
How many 4-digit numbers can be formed by the digits 2 4 6?
81 As there are no limits stated then you can have a number comprising a repeated single digit (such as 2222), two pairs of numbers (e.g. 2244) or three different numbers (such as 2462). The first digit can be one of any of the 3 numbers. The second digit can be one of any of the three numbers, as can the third digit and also the fourth. Then you can have 3 x 3 x 3 x 3 = 81 different 4-digit numbers using the three given numbers.
How many 4 digit numbers can be formed from the digits 12345?
The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5
How many 4-digit numbers can be formed from digits 123456789?
-123456785
How do you solve this 1 3 4 7 11?
Are you posting questions to which you already know the answers?Add the first two digits and get the third; add the second and third digits and get the fourth, add the third and fourth and get the fifth, etc. Next is 18.
How many different four-digit numbers can you make using the digits 2 4and 6?
First digit can be any of 3, second can be any of 3, as can third and fourth so your answer is 3 to the fourth ie 81
How many four digit numbers can be formed using the digits 5 8 and 9 with one pair of adjacent fives and no other repeated digits?
For every pair of 5s, the other two digits can be filled in 2 × 1 = 2 different ways. There are 3 positions the pair of 5s can sit (first and second, second and third, or third and fourth digits) → there are 3 × 2 = 6 different possible numbers (which are: 5589, 5598, 8559, 9558, 8955, 9855)
How many different numbers can be form by arranging the four digit 2013?
by multiplying how many digits there are to the fourth power and dividing that by 4 get 64. so u can possibly make 64 combinations... it think
How many 4 digit numbers have 6 as their second digit?
There are 9 digits that can be the first digit (1-9); for each of these there is 1 digit that can be the second digit (6); for each of these there are 10 digits that can be the third digit (0-9); for each of these there are 10 digits that can be the fourth digit (0-9). → number of numbers is 9 × 1 × 10 × 10 = 900 such numbers.
How many different 4 digit combinations can be made using the digits 0 to 9?
Well you have 10 possible numbers for the fist column(0-9) and 10 for the second, third, and fourth then you multiply those numbers. 10*10*10*10=10000 or 10^4=10,000. So there are 10,000 different combinations.
How many four digits numbers can be made with 1 4 5 9 without repetition?
To calculate the number of four-digit numbers that can be made using the digits 1, 4, 5, and 9 without repetition, we use the permutation formula. Since there are 4 digits to choose from for the first digit, 3 for the second, 2 for the third, and 1 for the fourth, the total number of permutations is 4 x 3 x 2 x 1 = 24. Therefore, there are 24 different four-digit numbers that can be formed using the digits 1, 4, 5, and 9 without repetition.
How many 4-digit numbers can be formed by the digits 2 4 6?
81 As there are no limits stated then you can have a number comprising a repeated single digit (such as 2222), two pairs of numbers (e.g. 2244) or three different numbers (such as 2462). The first digit can be one of any of the 3 numbers. The second digit can be one of any of the three numbers, as can the third digit and also the fourth. Then you can have 3 x 3 x 3 x 3 = 81 different 4-digit numbers using the three given numbers.
How many 4 digit numbers can be formed from the digits 12345?
The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5
If you create a password using 4 numbers 0 through 9 how many different combinations are there without using the same number twice?
This is a factorial problem. The first number can be any of ten digits, the second any of nine (because you can't repeat a digit), the third any of eight and the fourth any of the remaining 7 digits. 10x9x8x7=5040 combinations.
How many 4 digit numbers can be made out of the numbers 0-9 and what are they?
Assuming several things: Numbers can't start with zero; Repeated digits are allowed, then: First digit can be any of 9, Second digit can be any of 10, Third and fourth digits can each be any of 10; There are therefore 9 x 10 x 10 x 10 ie 9,000 possible answers, which by coincidence is the total of the numbers from 1000 to 9999. If howerver you did not intend to allow repeated digits then the first digit can be one of 9, the second also one of 9 (zero now allowable), the third can be one of 8 and the fourth one of 7, giving a total of 9 x 9 x 8 x 7 ie 4536 arrangements, some of which will contain the same four digits but in a different order eg 1234 and 1243. If you don't want this then divide by 24 and get a more manageable 189.
What is the product of the fourth and sixth prime numbers?
The product of the fourth and sixth prime numbers is 91.
How many 4 digit combination's can you get from numbers 1-20?
This isn't a very specific question ... by four digits, do you mean four numbers? Furthermore, are the digits or numbers you use allowed to repeat? Let's assume you meant numbers, not digits. In this case, for example, you could be trying to create a locker combination, and the numbers you can use are 1 through 20, and the combination has four "blank" spaces. 1. If the numbers can repeat, then each spot has 20 possibilities, so the total number of combinations is 20 for the first spot, 20 for the second, 20 for the third, and 20 for the fourth. That is, 20*20*20*20 = 160,000 combinations. 2. If the numbers cannot repeat, then the first spot can be taken up by any of 20 numbers, the second spot only by 19 (because you can't repeat the first number, whatever it was), the third spot by 18, and the fourth spot by any one of the 17 remaining numbers. So that makes 20*19*18*17 possibilities = 116,280 combinations. If this is not what you meant, and you wanted digits, remember that every number above and including 10 has two digits, so you'll want to factor that into your calculations.
