Since there are 10 numbers to fill 4 spots, each spot would have 10 choices for which number would fill it. We have (10)(10)(10)(10)=1000.
The number of permutations of a set of distinct objects is calculated using the factorial of the number of objects. For the numbers 10 through 14, there are 5 distinct numbers (10, 11, 12, 13, and 14). Therefore, the number of permutations is 5! (5 factorial), which equals 5 × 4 × 3 × 2 × 1 = 120.
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.
There are 90 two-digit numbers... starting with 10 and ending with 99.
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.
There are 9*10*9 = 810 such numbers.
In case of digits or numbers, when we say combination, since the order of the numbers matters, we really need to find the permutations. For example, 12 and 21 are two permutations and not combination. The permutation for 10 digits chosen 5 ways is 10*9*8*7*6, which is 30240. But how many are there in whole like four digit has 24.
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).
To calculate the number of different 4-digit combinations that can be made using numbers 0 through 9, we use the concept of permutations. Since repetition is allowed, we use the formula for permutations with repetition, which is n^r, where n is the number of options for each digit (10 in this case) and r is the number of digits (4 in this case). Therefore, the number of different 4-digit combinations that can be made using numbers 0 through 9 is 10^4, which equals 10,000 combinations.
The number of permutations of a set of distinct objects is calculated using the factorial of the number of objects. For the numbers 10 through 14, there are 5 distinct numbers (10, 11, 12, 13, and 14). Therefore, the number of permutations is 5! (5 factorial), which equals 5 × 4 × 3 × 2 × 1 = 120.
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.
the right answer is 10 cause there is o,1,2,3,4,5,6,7,8,9 so there 10 numbers in total which have 1 digit
There are 90 two-digit numbers... starting with 10 and ending with 99.
ten factorial = 10! = 3,628,800
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.
Number of 7 digit combinations out of the 10 one-digit numbers = 120.
72. (with the range of two digit numbers being from 10 to 99).
There are 9*10*9 = 810 such numbers.