16 times
The question is incorrect since there are 210 - 1 possible combinations. The digit 0 can be in the combination or out. That gives 2 ways. With each, the digit 1 can be in the combination or out - 2*2 = 22 ways. With each, the digit 2 can be in the combination or out = 23 ways. With each, the digit 3 can be in the combination or out = 24 ways. etc. So 210 ways in all except that one of them is the null combination. Now 210 = 1024 so there are only 1023 combinations. If, instead, you allow the digits to be used many times, there is no limit to the number of combinations.
Only 1. In a combination, the order of the numbers does not matter. So 102478 is the same combination as 104782 or 407812 etc.
6,720 combinations.
10x9x9x9
16 times
The question is incorrect since there are 210 - 1 possible combinations. The digit 0 can be in the combination or out. That gives 2 ways. With each, the digit 1 can be in the combination or out - 2*2 = 22 ways. With each, the digit 2 can be in the combination or out = 23 ways. With each, the digit 3 can be in the combination or out = 24 ways. etc. So 210 ways in all except that one of them is the null combination. Now 210 = 1024 so there are only 1023 combinations. If, instead, you allow the digits to be used many times, there is no limit to the number of combinations.
Only 1. In a combination, the order of the numbers does not matter. So 102478 is the same combination as 104782 or 407812 etc.
There is only one combination. There are many permutations, though.
6,720 combinations.
10x9x9x9
There are 126 different 5 digit combinations. Note that the combination 12345 is the same as the combination 45312.
You have a three digit number that can only use each digit once in combination. Therefore, a number like 456 is acceptable, but 455 or 101 or 222 is not acceptable. I think you can solve the problem in the following manner. There are 10 possible digits (0-9) that can be used in the first number. For the second number there are only 9 possible since you cannot repeat one of the first. For the third number there are only 8 possible since again you cannot repeat any of the first two. You multiply the possibilities for each digit to get our answer. 10 * 9 * 8 = 720 There are 720 possible combinations.
61
Using the eight digits, 1 - 8 ,-- There are 40,320 eight-digit permutations.-- There is 1 eight-digit combination.
Choose from 4 for the first number, 3 for the second number, and 2 for the third number. Therefore there are 4*3*2 = 24 three digit numbers can be formed from 3769 if each digit is used only once.
4 options for the first digit, 3 options for the second digit, 2 options for the third digit. Multiply the number of options together, and you find how many 3-digit numbers you can get.