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.
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
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%.
4284
There are 90 palindromes with 4 digits.The first digit can be any digit from the set {1,2,3,4,5,6,7,8,9}.With each choice of the first digit, the second can be any digit from the set {0,1,2,3,4,5,6,7,8,9}.That makes 9*10 = 90 permutations for the first two digits. These determine the palindrome since the third and fourth digits are the same as the second and first, respectively.