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.
First digit can be any of 3, second can be any of 3, as can third and fourth so your answer is 3 to the fourth ie 81
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.
2578 5287 5782 8572
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 only positive digits, and disregarding their order, 3 ways.
1
6 if all digits are different, 27 otherwise.
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