To form a three-digit number using the digits 0-9, the first digit cannot be 0 (as it would not be a three-digit number). Thus, the first digit can be any of the digits from 1 to 9 (9 options). The second and third digits can each be any digit from 0 to 9 (10 options each). Therefore, the total number of three-digit numbers is (9 \times 10 \times 10 = 900).
There are 76 such numbers. Eight more if you allow numbers to start with 0.
There are 12 such numbers.
-- The first digit can be any one of nine. (It can't be zero).-- The second digit can be any one of nine. (Anything except the one you used for the first digit.)-- The third digit can be any one of eight. (Anything except the two already used.)Total possibilities = (9 x 9 x 8) = 648
24 of them.
210
There are 76 such numbers. Eight more if you allow numbers to start with 0.
Assuming that numbers with a leading zero is not permitted, the answer is 9*9! = 3,265,920. If leading 0s were permitted, the answer would have been 3,628,800
id say, 0123456789. but depends how many digits you can have
There are 12 such numbers.
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).
999-111=888 888 3 digit numbers can be made with numbers between 1 - 9
-- The first digit can be any one of nine. (It can't be zero).-- The second digit can be any one of nine. (Anything except the one you used for the first digit.)-- The third digit can be any one of eight. (Anything except the two already used.)Total possibilities = (9 x 9 x 8) = 648
a lot
24 of them.
90000
2
210