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 9*10*9 = 810 such numbers.
720 (10*9*8)
30,25,15,20, or 10.
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
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).
3
There are 9*10*9 = 810 such numbers.
There are 900 three digit numbers. (99 - 1000) (# of possible numbers in the first position = 9) (# of possible numbers in the second position = 10) (# of possible numbers in the third position = 10) 9 *10 *10 = 900
720 (10*9*8)
30,25,15,20, or 10.
You can select 12 numbers for the first digit, 11 numbers for the second digit, and 10 numbers for the third digit; so 12*11*10 = 1320 sets of 3 numbers can be made out of 12 different numbers.