The question does NOT say that the three places must be different digits, so repetition is allowed. The first digit can't be zero, so there are 9 choices for it.
The second digit can be any one of the 10, including zero.
The third digit can also be any one of the 10, including zero. So the quantity of 3-digit numbers that can be written is (9 x 10 x 10) = 900. (This makes sense. The smallest number you can write with 3 digits is 100, and the largest is 999, which is a total of 900 different numbers.)
If repetition is not allowed, then: The first digit can't be zero, so there are 9 choices for it.
For each choice, the second digit can be any one of the remaining 9 (including zero).
For each choice, the third digit can be any one of the remaining 8 (including zero). So the quantity of 3-digit numbers that can be written without repetition or a leading zero is (9 x 9 x 8 ) = 648.
But the question doesn't ask "how many numbers can be made".
It asks "How many combinations can be made ?" When you say "combinations" in math, that ordinarily means that the order doesn't matter, i.e. 247 would be considered the same thing as 724. Granted, this is getting pretty picky, but that's exactly what math is all about, so we need to consider the possibility that the question was actually referring to combinations, and the order of the digits doesn't matter. If repetition is not allowed: The first digit can be any one of 10, including zero.
For each choice, the second digit can be any one of the remaining 9.
For each choice, the third digit can be any one of the remaining 8.
The total number of 3-digit groups is (10 x 9 x 8) = 720.
Each group of 3 digits can be arranged in (3 x 2) = 6 ways. So if the order doesn't matter, there are (720/6) = 120separate, distinct COMBINATIONS of 3 digits.
If the numbers are allowed to repeat, then there are six to the fourth power possible combinations, or 1296. If they are not allowed to repeat then there are only 360 combinations.
10 Combinations (if order doesn't matter). 3,628,800 Possiblilities (if order matters).
There are just 5,005.There are 10C6 = 210 combinations with 6 different numbers.There are 10C5 = 252 combinations with 5 different numbers. One of these needs to be duplicated and the one to be duplicated can be chosen in 5 ways so there are 1260 of these.There are 10C4 = 210 combinations with 4 different numbers. Either one of them appears three times or two are doubled. That can be done in 10 ways so there are 2,100 of these.There are 10C3 = 120 combinations with 3 different numbers. Either one of them appears 4 times, or one twice and another three times, or all three appear twice. That can be done in 10 ways so there are 1,200 of these.There are 10C2 = 45 combinations with 2 different numbers. Either one of them appears 5 times, or one twice and another four times, or both three times. That can be done in 5 ways so there are 225 of these.There are 10C1 = 10 combinations with just 1 number repeated six times.Add these together.
720 is the number of ways to combine three known letters and three known numbers.For example, the letters A, B & C and the numbers 1, 2 & 3. The total combinations of these 6 characters is:(6 options)*(5 options)*(4)*(3)*(2)*(1) = 720.However, if the three numbers and three letters are unknown and any number or letter is possible, and repeated numbers or letters are acceptable (such as with a license plate), then the total possibilities for each "space" are multiplied together:(26 possibilities)*(26 possibilities)*(26 possibilities)*(10 possibilities)*(10 " ")*(10 " ") = 17,576,000 combinations.That is, there are 26 letters in the alphabet and 10 numbers (0 thru 9).This is assuming that three of the six spaces spaces are reserved for letters and three spaces are reservedfor numbers.If the combination can be any three letters and any three numbers where different combinations are made by changing whether each space contains a number or a letter, then the answer becomes a product and sum of different choose functions and is much more complicated...
The only numbers that can be made with ONLY 5 and 6 are 56 and 65. That is because 5 and 6 are only 2 numbers. If you wanted a three digit number, you would have to use double of one number. You could say these numbers: 556 566 565 656 665 Other than that, I do not know.
6
32C3 = 4960
45
Six combinations: 123, 132, 213, 231, 312, 321
Assuming you are treating each number as a number and not as an individual unit, the numbers you can make from these digits are 899, 989 and 998.
017, 071, 107, 170, 701, 710. 6 combinations
10
9*9*9 = 729 using the digits 1 to 9 and 2*9 using 10 and another digit. 749 in all.
We notice that you said "combinations of three digits". You didn't say "3-digit numbers".There are (10 x 10 x 10) = 1000 combinations of three digits.The only combinations of 3 digits that form numbers less than 15 are 000, 001, 002, . . .012, 013, and 014. There are 15 of those.So the remaining (1000 - 15) = 985 are all larger than 15.
-8
There are 720 combinations if you use each of the digits only once per combination.
i guess the answer is 54. A group of three digits can be selected from the 5 digits in 9 ways and each group can be arranged in 3! (i.e., 6) ways respectively and hence the answer is 6*9=54