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.
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.
14: Quarters Dimes Nickels 0 0 13 0 1 11 0 2 9 0 3 7 0 4 5 0 5 3 0 6 1 1 0 8 1 1 6 1 2 4 1 3 2 1 4 0 2 0 3 2 1 1
1 of 0 elements 5 of 1 elements 10 of 2 element 10 of 3 elements 5 of 4 elements 1 of 5 elements A total of 25 = 32 combinations.
That depends on . . . -- how many digits you want in each combination -- whether it's allowed to repeat the same digit -- whether the order of the digits in the combination matters, i.e. whether you actually want combinations or permutations. Each choice has a different answer.
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
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
24
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.
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.
Precisely 1. For ANY number n, the number of combinations of n of those items is always 1.
14: Quarters Dimes Nickels 0 0 13 0 1 11 0 2 9 0 3 7 0 4 5 0 5 3 0 6 1 1 0 8 1 1 6 1 2 4 1 3 2 1 4 0 2 0 3 2 1 1
The number of combinations is 10C4 = 10*9*8*7/(4*3*2*1) = 210