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
With a total of 90,000 five digit numbers, we can conclude that there is 45,000 EVEN five digit numbers.
6 x 6 x 6 x 3 = 648 6 because the first digit can be any of the numbers 6 again, because the second digit can be any of the numbers 6, again, because the third digit can be any of the numbers 3, because the fourth/last digit can only be 2, 4, or 6
There are 9 1-digit numbers and 16-2 digit numbers. So a 5 digit combination is obtained as:Five 1-digit numbers and no 2-digit numbers: 126 combinationsThree 1-digit numbers and one 2-digit number: 1344 combinationsOne 1-digit numbers and two 2-digit numbers: 1080 combinationsThat makes a total of 2550 combinations. This scheme does not differentiate between {13, 24, 5} and {1, 2, 3, 4, 5}. Adjusting for that would complicate the calculation considerably and reduce the number of combinations.
There are 9 choices for the first digit (1,2,3,4,5,6,7,8,9) No zero because numbers don't start with zeroes. 10 choices for the second digit (0,1,2,3,4,5,6,7,8,9). 10 choices for the third digit (0,1,2,3,4,5,6,7,8,9) Okay now 9x10x10=900. Answer:900
first digit time second digit and second digit times first digit then repeat
Write them as decimals, and compare. If the first digit of two numbers is equal, compare the second digit; if the second digit is equal, compare the third digit, etc.
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.
90
The second digit can be 2,4,6, or 8. There are 4 ways of choosing the second digit. The first digit can be chosen from the remaining 8 digits. 8x4 = 32 numbers. Here, it is assumed that the numbers do not repeat
The first digit can have 5 possible numbers, the second digit can have 4, the third 3, the fourth 2. 5
2478 is MMCDLXXVIII
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.
1000... 0600-0699 1600-1699 and so on.
The first digit has 4 choices for its digit. The second digit has 6 choices and the third has 3. The solution would simply be 4*6*3=72 three digit numbers.
In such cases, you should compare one digit at a time, from left to right, until you find a digit that is different in the two numbers. That is, compare the first digit (after the decimal period) with the first digit, the second digit with the second digit, etc.
You can select 9 numbers for the first digit, 8 numbers for the second digit, and 7numbers for the third digit; so 504 (e.g. 9*8*7) different three digit numbers can be written using the digits 1 through 9.