By using the 3 digits of a number we can form 3 different two digit numbers.
3C2 = 3!/[(3 - 2)!2!] = 3!/(1!2!) = (3 x 2!)/2! = 3
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Assuming that the first digit of the 4 digit number cannot be 0, then there are 9 possible digits for the first of the four. Also assuming that each digit does not need to be unique, then the next three digits of the four can have 10 possible for each. That results in 9x10x10x10 = 9000 possible 4 digit numbers. If, however, you can not use the same number twice in completing the 4 digit number, and the first digit cannot be 0, then the result is 9x9x8x7 = 4536 possible 4 digit numbers. If the 4 digit number can start with 0, then there are 10,000 possible 4 digit numbers. If the 4 digit number can start with 0, and you cannot use any number twice, then the result is 10x9x8x7 = 5040 possilbe 4 digit numbers.
then there would only be a 1. and there would be no 2,3,4,5,6,7,8,9 digit numbers onlya one digit number. and there would be only 1,2,3,4,5,6,7,8,9,numbers to use not 10, 20,30,40 ,50 ,60,70,80,90,100 because they need zeros
To determine the number of 3-digit numbers that are multiples of 5, we need to find the first and last 3-digit multiples of 5. The first 3-digit multiple of 5 is 100, and the last 3-digit multiple of 5 is 995. To find the total number of such multiples, we can use the formula (Last - First) / 5 + 1 = (995 - 100) / 5 + 1 = 180. Therefore, there are 180 3-digit numbers that are multiples of 5.
In the decimal number system, the highest valued digit is 9. The highest digit that ever appears in any one 'place' of a number is one less than the 'base' of the number. The numbers that everyone is most familiar with ... the numbers you see around you every day ... are numbers written in the 'decimal' system, using the 'base' of 10. So the highest digit in any one place is 9. 'Binary' numbers ... the form most used to represent numbers inside digital circuits and computers ... are constructed in base 2. So the highest digit in any one place is 1, and each of these numbers is just a string of 1's and zeros. Digits can be even higher than 9 in number systems that use other bases. For example, the hexadecimal system (often used in computer science to represent binary numbers) is base 16, so in that case the highest valued digit is "F" which has a value equivalent to 15 in a decimal representation. As an example, the number "FA" hexadecimal, has decimal value 15*16 + 10 = 250.
To calculate the number of 3-digit combinations that can be made from the numbers 1-9, we can use the formula for permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the total number of options (10 in this case) and r is the number of digits in each combination (3 in this case). Therefore, the total number of 3-digit combinations that can be made from the numbers 1-9 is 10^3 = 1000.