21 ways assuming no leading zeros, that is the smallest possible number with three digits is 100
Otherwise if leading zeros are allowed, it is 28 ways.
For three digit numbers the first digit can be 1-6. Leaving the remaining two digits to be 6-value_of_first_digit.
For first digit 6, the remain two digits sum to 0 which means they can only be 00, ie the number 600 - 1 number
For first digit 5, the remain two digits sum to 1 which means they can be 01 or 10, ie the numbers 501, 510 - 2 numbers
For first digit 4, the remain two digits sum to 2 which means they can be 02. 11, 20, ie the numbers 402, 411, 420 - 3 numbers
etc
It can be seen that the number of ways of making a two digit number (with leading zeros) sum to a number less than 10 is the required sum plus 1. So for the remaining first digits, the number of ways are: 3 -> 4 ways, 2 -> 5 ways, 1 -> 6 ways.
Thus the total number of ways is 1 + 2 + .. + 6 = 21
If leading zeros are permitted, so that, for example, 060 (60) and 006 (6) are considered as three digits numbers, then there are a further 7 ways with a first digit of 0, making a total 28 ways.
450.
None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits. None, because you do not have three even digits.
The final digit must be a 2 to form an even number. The first digit may be any of three remaining digits (1, 3, 5) while the second digit may be any of the two remaining digits. All together, that makes 3*2 = 6 distinct even numbers.
1000there aren't even 1000 three-digitnumbers...There are the digits 1 through 9 for the first digit. Then, we have 0 through 9 for the second digit - excluding the first digit. For the third digit, we have 0 through 9, excluding the two previous digits for a total of 9*9*8 =81*8 =648 three digit numbers with distinct digits.EDIT- ... to be a little more specific, if your talking about a 3-digit password or unlock code its would be 1000. _-_
The last digit must be odd: three possibilities. The other digits can be anything else, as long as they're different: five times four times three possibilities. Thus, the answer is 3 x 5 x 4 x 3 = 180.
27 three digit numbers from the digits 3, 5, 7 including repetitions.
There are 900 of them.
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
48 of them if digits may not be repeated. 100 if they can.
If the digits can be used more than once, then 900. If not, then 648.
Forming three three digit numbers that use the numbers 1-9 without repeating, the highest product possible is 611,721,516. This is formed from the numbers 941, 852, and 763.
If the digits may not be repeated, there are 6x5x4x3x2x1 = 720 numbers, If digits can be repeated, the answer is 6x6x6x6x6x6 = 46,656 numbers If zero is ong the available digits but is excluded from the first place. 5x6x6x6x6x6=38,880 numbers
5 x 10 x 5 = 250 different numbers, assuming there is no limit to each digits' use.
There are 900 three-digit numbers.
Using the digits of 1345678, there are 210 three digit numbers in which no digit is repeated.
24+24+12+4=64.
757