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In order for the number to be even, the '2' must be the last digit. So we only need to
figure out how many ways there are to arrange the 1, 3, and 5 in the first 3 places.
The first place can be any one of 3 digits. For each of those ...
The second place can be either one of the 2 remaining digits.
And then the third place has to be the only digit that's left.
So there are (3 x 2) = 6 ways to arrange 1, 3, and 5 in the first 3 places.
That means only 6 even 4-digit numbers can be made from the digits 1, 2, 3, and 5.
And here they are:
1 3 5 2
1 5 3 2
3 1 5 2
3 5 1 2
5 1 3 2
5 3 1 2
What else can I help you with?
What is the plural form for digits?
The plural form is digits; the singular form is digit.
How many different three digit numbers can you form using the digits 2 5 and 7?
6 possible 3 digit combonations
What is the word form for each number and the value of the underlined digit?
Since there are no numbers and no underlined digits, there cannot be any answer!
How many 4 digit numbers can you form from the digits 1 2 3 4 5 where no digit is used twice?
5*4*3*2 = 120
How many 3 digit numbers can be made from 0123456789?
To form a three-digit number using the digits 0-9, the first digit cannot be 0 (as it would not be a three-digit number). Thus, the first digit can be any of the digits from 1 to 9 (9 options). The second and third digits can each be any digit from 0 to 9 (10 options each). Therefore, the total number of three-digit numbers is (9 \times 10 \times 10 = 900).
How many even 3-digit numbers can be formed from the digits 1 2 3 and 5 if each digit can be used only once?
The final digit must be a 2 to form an even number. The first digit may be any of three remaining digits (1, 3, 5) while the second digit may be any of the two remaining digits. All together, that makes 3*2 = 6 distinct even numbers.
How many two digit numbers can you form using the digits 1235679 with repetition allowed?
You have seven different digits (symbols) to choose from, so you can form seven different one digit numbers and 7×7=72=49 different two digit numbers.
How many 2 digit numbers digits can be written in accending form?
90 of them.
How many different numbers can you form if no digit is repeated?
The answer will depend on how many digits you have at your disposal. Working with the 10 digits, and barring leading zeros, there are 8,877,690 numbers.
How many 6 digit numbers can be formed form the digits 123456 if no digits can be repeated?
6*5*4*3*2*1=720 possible numbers
What is the plural form for digits?
The plural form is digits; the singular form is digit.
How many different three digit numbers can you form using the digits 2 5 and 7?
6 possible 3 digit combonations
How many six digit numbers can be formed using the digits 0123456789 Example equals 123456 or 098765?
The number of six digit numbers that you can make from ten different digits ifrepetitions of same digit on the six digit number is allowed is 1 000 000 numbers(including number 000 000).If no repetitions of the the same digit are allowed then you have:10P6 = 10!/(10-6)! = 151 200 different six digit numbers(six digit permutations form 10 different digits).
What is the word form for each number and the value of the underlined digit?
Since there are no numbers and no underlined digits, there cannot be any answer!
How many different three digit numbers can you form using the digits 1 2 3 5 6 7 and 9 without repetition?
There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.
How many 3 digit numbers can be formed?
It depends upon whether the numbers can be used more than once. If the numbers can be used more than once, then there are 1,000 possible combinations; if not, then there are 720 possible combinations. ========== Assuming you are talking about integers you can calculate it this way: you can have any one of 9 digits for the first digit (zero is excluded because that would make it only a 2 digit number) You can have any one of 10 digits for the second and any one of 10 digits for the third digit. That gives you 9x10x10 = 900 different combinations for 3 digit numbers (not 1000). If you can include both negative and positive numbers as different numbers you get twice as many (2x900=18000). If you can count decimal numbers as 3 digit numbers (i.e. not restricted to integers) you can have 900 integers, 900 with the form xx.x, 1000 with the form x.xx (if zero is permitted to be the first digit and count as one of the 3 digits) or 900 (if a leading zero is NOT counted as one of the 3 digits). If a leading zero is NOT counted as one of the 3 digits, you could also have 1000 numbers of the form 0.xxx or just .xxx
How many 4 digit numbers can you form from the digits 1 2 3 4 5 where no digit is used twice?
5*4*3*2 = 120
