The highest number you can make out of the digits 2 4 0 and 1 is 4210.
29 x 2 + 17 x 3 - 9 = 58 + 51- 9 = 100
You can make: 1 combination containing 0 digits, 7 combinations containing 1 digits, 21 combinations containing 2 digits, 35 combinations containing 3 digits, 35 combinations containing 4 digits, 21 combinations containing 5 digits, 7 combinations containing 6 digits, and 1 combinations containing 7 digits. That makes 2^7 = 128 in all.
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
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.
The highest number you can make out of the digits 2 4 0 and 1 is 4210.
3 * (2 - 1)
29 x 2 + 17 x 3 - 9 = 58 + 51- 9 = 100
2*3*3*3*3 = 162
20 + 5 + 2 and two 1 cents will make 29 Euro cents.20 + 5 + 2 and two 1 cents will make 29 Euro cents.20 + 5 + 2 and two 1 cents will make 29 Euro cents.20 + 5 + 2 and two 1 cents will make 29 Euro cents.
33 = 27 with repetition, 3! = 3*2*1 = 6 without repetition.
First number is 20 and last is 99 so 20 = 1 then 21 = 2 then 29 = 10 So from 29 take 19 to make 10 then apply to 99 as in 99 - 19 = answer 80
You can make: 1 combination containing 0 digits, 7 combinations containing 1 digits, 21 combinations containing 2 digits, 35 combinations containing 3 digits, 35 combinations containing 4 digits, 21 combinations containing 5 digits, 7 combinations containing 6 digits, and 1 combinations containing 7 digits. That makes 2^7 = 128 in all.
32 - 4 + 1 =29
Assuming the digits cannot be repeated, there are 7 combinations with 1 digit, 21 combinations with 2 digits, 35 combinations with 3 digits, 35 combinations with 4 digits, 21 combinations with 5 digits, 7 combinations with 6 digits and 1 combinations with 7 digits. That makes a total of 2^7 - 1 = 127: too many for me to list. If digits can be repeated, there are infinitely many combinations.
1023
1 + 1 = 2 The sum of the digits is therefore 2.