11 + 4, 12 + 3, 13 + 2, 14 + 1, 15 + 0,
15 - 0, 16 - 1, 17 - 2, 18 - 3, 19 - 4,
20 - 5, 21 - 6, 22 - 7, 23 - 8, 24 - 9,
1*6 + 9, 1*7 + 8, 1*8 + 7, 1*9 + 6,
2*3 + 9, 2*4 + 7, 2*5 + 5, 2*6 + 3, 2*7 + 1, 2*8 - 1, 2*9 - 3
3*2 + 9, 3*3 + 6, 3*4 + 4, 3*5 + 0, 3*6 - 3, 3*7 - 6, 3*8 - 9
4*2 + 7, 4*3 + 3, 4*4 - 1, 4*5 - 5, 4*6 - 9
5*2 + 5, 5*3 + 0, 5*4 - 5
6*1 + 9, 6*2 + 3, 6*3 - 3, 6*4 - 9
7*1 + 8, 7*2 + 1, 7*3 - 6
8*1 + 7, 8*2 - 1, 8*3-9
9*1 + 6, 9*2 - 3
1*3*5
Is that enough?
If you want 4-digit numbers, there are 24 of them.
9
How many permutations of 3 different digits are there, chosen from the ten digits 0 to 9 inclusive?
600 of them.600 of them.600 of them.600 of them.
Well honey, you've got 4 digits there, so you can form 4! (4 factorial) which is 24 numbers. That's right, you can make 24 different combinations with those digits. Math can be fun when you've got some sassy numbers to play with!
89,999 different numbers i guess
Yuo can make only one combination of 30 digits using 30 digits.
111, 121, 212, 222
If the digits are all different then 18. Otherwise, 192.
It is possible to create infinitely many numbers, of infinitely many different lengths, using the digits of the given number. Using each of the digits, and only once, there are 5! = 120 different permutations.
Using the digits 1, 3, and 5 exactly once, you can create different 3-digit numbers by permuting these digits. The number of permutations of 3 distinct digits is calculated as 3! (3 factorial), which equals 6. Therefore, the different numbers you can create are: 135, 153, 315, 351, 513, and 531. Thus, there are 6 different numbers that can be formed.
6 if all digits are different, 27 otherwise.
Using only positive digits, and disregarding their order, 3 ways.
1
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.
If you want 4-digit numbers, there are 24 of them.
0