Think of either using or not using the numbers 1 and 4 for each digit. For example, let ____ ____ _____ ____ be the 4 digit number, now 1 can either go in to the first spot or 4 can. So there are 2 choices for that spot, and similarly two for the next spot and the next and the last. So there are 2^4 =16 combinations of 1 and 4 if we make 4 digit numbers. (this uses the multiplication rule) To help a little more to see it, look at just the two digit number, It should have 2^2=4 choices by the logic above. Here they are: 11,14,41,44 and that's it!. See how we had two choices for each digit?
If each number can be used only once then there are just four: 2345, 2346, 2456 and 3456. There are no others because the order of the digits does not matter in combinations.
If the numbers contain zeros, the total number of combinations is 10,000. You can work this out easily logically: For ten single-digit numbers (0,1,2,3,4,5,6,7,8,9) then there are 10 possible 'combinations' For numbers with 2 digits then for each possible digit in the 10s column (e.g. in the 20s range) there are another 10 possible combinations (20,21,22,23,24,25,26,27, 28,29). As there are 10 possible ranges (single digits, teens, twenties, thirties etc) there will be 10 X 10 or 100 possible combinations. using the same logic, for three digits, there will be 10 X 10 X 10 or 1000 digits. And for 4 digits there will be 10 x 10 x 10 x 10 = 10,000 possible combinations. So for a number, say, with x digits, the total number of combinations of those digits will be 10 x 10 x 10..... etc with x numbers of 10s in the calculation. You can find out the number of combinations of any set of letters or numbers in the same way. as an example, to find out, say, the possible combinations of letters in the alphabet of 26 letters, then using the same method this can be given as 26 x 26 x 26 x 26............. with 26 '26's' in a row multiplied together. This gives the staggering amount of approximately 615612 followed by 31 zeros.
If the digits are all different then 18. Otherwise, 192.
how many numbers exactly have 4 digits ? 8900, 8999, 9000, 9999
-- If you're allowed to repeat digits in one combination,then there are 104 = 10,000 possibilities.-- If all four of the digits must be different, then there are(10 x 9 x 8 x 7) = 5,040 possibilities.
45
10
4!=4x3x2x1=24
If each number can be used only once then there are just four: 2345, 2346, 2456 and 3456. There are no others because the order of the digits does not matter in combinations.
You can get only four combinations: They are: 11, 118, 119 and 1189. In a combination, the order of the digits does not matter.
3 x 3 x 3 x 3 = 81 combinations
Not repeating, it is 7*6*5*4 which is 840 ways ---- There are 7 choices for each of four digits, right? 74 = 2401
If the numbers contain zeros, the total number of combinations is 10,000. You can work this out easily logically: For ten single-digit numbers (0,1,2,3,4,5,6,7,8,9) then there are 10 possible 'combinations' For numbers with 2 digits then for each possible digit in the 10s column (e.g. in the 20s range) there are another 10 possible combinations (20,21,22,23,24,25,26,27, 28,29). As there are 10 possible ranges (single digits, teens, twenties, thirties etc) there will be 10 X 10 or 100 possible combinations. using the same logic, for three digits, there will be 10 X 10 X 10 or 1000 digits. And for 4 digits there will be 10 x 10 x 10 x 10 = 10,000 possible combinations. So for a number, say, with x digits, the total number of combinations of those digits will be 10 x 10 x 10..... etc with x numbers of 10s in the calculation. You can find out the number of combinations of any set of letters or numbers in the same way. as an example, to find out, say, the possible combinations of letters in the alphabet of 26 letters, then using the same method this can be given as 26 x 26 x 26 x 26............. with 26 '26's' in a row multiplied together. This gives the staggering amount of approximately 615612 followed by 31 zeros.
If the digits are all different then 18. Otherwise, 192.
There is 1 combination of all ten numbers, 10 combinations of one number and of nine numbers, 45 combinations of two or eight numbers, 120 combinations of three or seven numbers, 210 combinations of four or six numbers and 252 combinations of five numbers. That is 1023 = 210 - 1 in total.
If you use them only once each, you can make 15 combinations. 1 with all four digits, 4 with 3 digits, 6 with 2 digits, and 4 with 1 digit. There is also a combination containing no digits making 16 = 24 combinations from 4 elements.
It is the number of combinations of four numbers where the number of available digits starts at 10 and reduces by 1 each time.