Calculate how many ways to get change of 78 I been asked to calculate how many ways there are to get change of 78 cents with the coins of 25,10,5,1.
I been able to write this down:
$25a + 10b + 5c + d = 78$
But I do not know how to continue. Can you help me please?
 A: It's the same as the coefficient of $x^{78}$ in $$(x^3+x^8+x^{13}+\cdots+x^{78})(1+x^5+x^{10}+\cdots+x^{75})(1+x^{10}+x^{20}+\cdots+x^{70})(1+x^{25}+x^{50}+x^{75})$$ That doesn't help you much by hand, but it gives you something concrete to enter into a CAS if that helps.
A: There may be a faster, more clever method, but I would do this by nested casework, on the number of coins of each size.

Case 1: No quarters
$\hspace{1cm}$ Case 1.1: No dimes
$\hspace{1cm}$$\hspace{1cm}$ Our final set of coins is fully determined by how many nickels we have. There are $16$ $\hspace{1cm}$ $\hspace{1cm}$ choices for the number of nickels $\left(0, 5, 10, \dots 75\right)$.
$\hspace{1cm}$ Case 1.2: One dime
$\hspace{1cm}$$\hspace{1cm}$ Similarly, we have $14$ choices for the number of nickels.
$\hspace{1cm}$ Case 1.3: Two dimes
$\hspace{1cm}$$\hspace{1cm}$ $12$ in this case. You should notice the pattern, and hopefully the reason for it.
$\hspace{1cm}$ Case 1.4: Three dimes
$\hspace{1cm}$$\hspace{1cm}$ $10$
$\hspace{1cm}$ Case 1.5: Four dimes
$\hspace{1cm}$$\hspace{1cm}$ $8$
$\hspace{1cm}$ Case 1.6: Five dimes
$\hspace{1cm}$$\hspace{1cm}$ $6$
$\hspace{1cm}$ Case 1.7: Six dimes
$\hspace{1cm}$$\hspace{1cm}$ $4$
$\hspace{1cm}$ Case 1.8: Seven dimes
$\hspace{1cm}$$\hspace{1cm}$ $2$
Case 2: One quarter
$\hspace{1cm}$ Case 2.1: No dimes
$\hspace{1cm}$$\hspace{1cm}$ Here we have $11$ choices for the number of nickels used. The corresponding cent values $\hspace{2.1cm}$ are $0$, $5$, $\dots 50$
$\hspace{1cm}$ Case 2.2: One dime
$\hspace{1cm}$$\hspace{1cm}$ We have a pattern again. $9$ in this case.
$\hspace{1cm}$ Case 2.3: Two dimes
$\hspace{1cm}$$\hspace{1cm}$ $7$
$\hspace{1cm}$ Case 2.4: Three dimes
$\hspace{1cm}$$\hspace{1cm}$ $5$
$\hspace{1cm}$ Case 2.5: Four dimes
$\hspace{1cm}$$\hspace{1cm}$ $3$
$\hspace{1cm}$ Case 2.6: Five dimes
$\hspace{1cm}$$\hspace{1cm}$ $1$ choice - no nickels
Case 3: Two quarters
$\hspace{1cm}$ Case 2.1: No dimes
$\hspace{1cm}$$\hspace{1cm}$ We have $28$ cents left to spend. There are $6$ possibilities for the number of nickels ($0$ to  $\hspace{1cm}$ $\hspace{1.1cm}$ $5$ nickels).
$\hspace{1cm}$ Case 2.2: One dime
$\hspace{1cm}$$\hspace{1cm}$ 4
$\hspace{1cm}$ Case 2.3: Two dimes
$\hspace{1cm}$$\hspace{1cm}$ 2
Case 4: Three quarters
$\hspace{1cm}$ There is only one way to complete this: three pennies.

Our final answer is the sum of the number of possibilities in each case. This is $$16 + 14 + 12 + 10 + 8 + 6 + 4 + 2 + 11 + 9 + 7 + 5 + 3 + 1 + 6 + 4 + 2 + 1 = \boxed{121}$$
A: Consider the product in alex.jordan's answer. The product of the last two factors is
$$1 + x^{10} + x^{20} + x^{25} + x^{30} + x^{35} + x^{40} + x^{45} + 2 x^{50} + x^{55}
+ 2 x^{60} +  x^{65} + 2 x^{70} + 2 x^{75} + \ldots)$$
The product of the first two is
$$ 1 + \ldots + x^4 + 2 (x^5 + \ldots + x^9) + 3 (x^{10}+\ldots+x^{14}) + \ldots
+ 16 (x^{75}+\ldots+x^{79})+\ldots$$
So the coefficient of $x^{78}$ is 
$$ \eqalign{ 16 & \text{ for } 1 \times 16 x^{78}\cr
+ 14 & \text{ for } x^{10} \times 14 x^{68}\cr
+ 12 & \text{ for } x^{20} \times 12 x^{58}\cr
+ \ldots\cr
+ 2 & \text{ for } 2x^{75} \times x^3}$$ 
