There are a lot of possibilities. The second digit can be 2 through 6, the third digit can be 3 through 7 as long as it is larger than the second digit. What we have so far: 1 _ _ 89
The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5
Oh, dude, let me break it down for you. So, you've got 8 numbers to choose from for each digit, and you're picking 4 digits in total. That means you have 8 choices for the first digit, 8 choices for the second digit, 8 choices for the third digit, and 8 choices for the fourth digit. Multiply all those together and you get... well, I'll let you do the math.
To calculate the number of 4-digit numbers that can be formed using the digits 1, 3, 5, 7, and 9 without repetition, we use the permutation formula. Since there are 5 choices for the first digit, 4 choices for the second digit (as one digit has been used), 3 choices for the third digit, and 2 choices for the fourth digit, the total number of 4-digit numbers that can be formed is 5 x 4 x 3 x 2 = 120.
Just compare the digits one by one: compare the first digit after the decimal point with the first digit of the other number, the second digit with the second digit, etc., until you find a digit that is different.
The fourth number in the decimal 5.248318 is 3. This is determined by counting the digits from the left, starting with the digit before the decimal point. In this case, the sequence of digits is 5 (first), 2 (second), 4 (third), and 3 (fourth).
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.
Not necessarily. Consider 444. The digits are not different. The first and second digits are not multiples of 3 The first digit is not greater than the second digit. In spite of all that, 444 is a 3-digit number
The first and last, and the second and third.
There are a lot of possibilities. The second digit can be 2 through 6, the third digit can be 3 through 7 as long as it is larger than the second digit. What we have so far: 1 _ _ 89
27
The first digit can be any one of 10 digits. For each of those . . .The second digit can be any one of the remaining 9 digits. For each of those . . .The third digit can be any one of the remaining 8 digits. For each of those . . .The fourth digit can be any one of the remaining 7 digits.The total number of possibilities is (10 x 9 x 8 x 7) = 5,040
1155
Alternate digits refer to the digits in a number that are separated by at least one other digit. For example, in the number 123456, the alternate digits would be 1, 3, and 5 (the first, third, and fifth digits) or 2, 4, and 6 (the second, fourth, and sixth digits). This concept is often used in mathematical problems and patterns to analyze specific sequences or properties of numbers.
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
1
The first digit can be any one of 9 digits (anything but zero). For each of those ...The second digit can be any one of 9 digits (anything but the previous one). For each of those ...The third digit can be any one of 9 digits (anything but the previous one). For each of those ...The fourth digit can be any one of 9 digits (anything but the previous one). For each of those ...The fifth digit can be any one of 9 digits (anything but the previous one).So the total number of possible arrangements is (9 x 9 x 9 x 9 x 9) = 59,049out of a total of 99,999 different 5-digit numbers, or 59.05%.