If you are looking for how many different ways to can arrange the ten digits (using each one exactly one time), there are 3,628,800 ways. You can get this by assuming :
1. There are ten choices for the first digit. Once you have picked that one...
2. There are nine remaining digits that you can choose for the second one. For two digits, then, there are 90 possibilities.
3. For each subsequent digit, multiply by the remaining number of possible choices. So:
10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1.
9.
There are 43 combinations of various quantities of quarters (0, 1 or 2), dimes (0 to 5), nickels (0 to 10) and pennies (2 to 52) that make 52 cents.
In each combination, you can have 0,1,2 or 3 pennies; you can also have 0 or 1 nickels, 0 or 1 dimes and 0 or 1 quarters. That is, 4 ways of picking the number of pennies, 2 ways of picking how many nickels, 2 for dimes and 2 for quarters. In all that makes 4*2*2*2 = 32 combinations. BUT, one of these combinations is where you have 0 pennies, 0 nickels, 0 diems and 0 quarters. Conventionally, a null combination such as this is omitted and so you have 31 valid combinations.
16 1 combination of all 4 4 combinations of 3 6 combinations of 2 4 combinations of 1 1 combination of 0
You Could Only Do It Ten (10) Combination :) :D
1
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.
9
You can make 5 combinations of 1 number, 10 combinations of 2 numbers, 10 combinations of 3 numbers, 5 combinations of 4 numbers, and 1 combinations of 5 number. 31 in all.
24
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.
Their is 25 combinations