Ways to add a number using just 1's, 5's, 10's, 25's, 50's

Question

given the set $\{1, 5, 10, 25, 50\}$, in how many ways, can you combine this numbers to get a specific number. For example, 11 can be shaped as $1\cdot11$, or $5\cdot112 + 1\cdot111$, or $10\cdot111 + 1$, or $5\cdot111 + 1\cdot116$. Then, for $11$, the answer is $4$. Suppose that for $0$, the answer is one. I think this is computation problem, but also a counting problem. Suppose numbers are cents, so the unique way to get a number is by adding.

Answer 1

N = new formed number.

Case 1: N = ab + cd + e: choose a number from the 5 numbers, there are 5 choices. After that there are 4 left. Chose a pair from 4 numbers. There are C(4,2) = 6 ways. So there are 5*6 = 30 new numbers formed this way.

Case 2: N = a + b + c + d + e: There is only one way: 1

Case 3: N = abcd + e: Choose a number out of 5 numbers. 5 choices.

Case 4; N = abc + de: Chose a pair from 5 numbers: there are C(5,2) = 10 choices.

Case 5: N = ab + c + d + e: Choose a pair from 5 numbers: there are C(5,2) = 10 choices.

Case 6: N = a(b + c + d + e): Choose a number from 5 numbers: there are 5 choices.

Case 7: N = (a + b)(c + d + e): choose a pair from 5 numbers. there are C(5,2) = 10 choices.

Case 8: N = ab(c + d + e): choose a pair from 5 numbers: 10 choices.

Case 9: N = abc(d + e): choose a pair from 5 numbers: 10 choices.

So by adding these up we get the total of newly formed numbers.

Answer 2

Generating functions are your friend for this. If you are looking to express $n$, you want the coefficient of $x^n$ in $(1+x+x^2+x^3+\dots)(1+x^5+x^{10}+\dots)(\dots)=\frac 1{1-x}\frac 1{1-x^5}(\dots)$