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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