I'm trying to understand the following problem.
A class collects 50 dollars to buy flowers for a classmate who is in the hospital. Roses cost 3 dollars each, and carnations cost 2 dollars each. No other flowers are to be used. How many different bouquets can be purchased for exactly 5o dollars?
What I"m confused about: what is this question even saying? Is there a specific amount of flowers in the bouquet or something? I don't understand what I'm supposed to do.
For this problem try thinking that roses as $-3r$ where each rose $r$ takes away $3$ dollars $-3r$. Do the same idea with the carnations, call the carnations $c$ and they cost $2$ dollars, $2c$.
Now your equation will look like: $50 = 3r+2c$ because you start out with $50$ dollars.
Now, we will solve for one of the variables. Let's solve for $c$.
$$50 = 3r+2c \\ 50 - 3r = 2c \\ \frac{50-3r}{2} = c$$
This final equation is equivalent to $y = \frac{50-3x}{2}$. Now you can either plug in numbers for $x$ and $y$ or plug it into a graphing calculator and see what your answers are. Note, that $r$ and $c$ I am assuming they are integers.
Consider the first quadrant (including the nonnegative axes). Let the $x$-coordinate represent the number of roses in a bouquet and the $y$-coordinate represent the number of carnations in a bouquet.
For each point in this quadrant, we can find the cost of the corresponding bouquet. How many points are there where the corresponding bouquet costs exactly 50 dollars? These few points represent the "different bouquets [that] can be purchased for exactly 50 dollars"
A slow method for solution: Since roses cost \$3, we can have at most $50/3 = 16.\overline{6}$ roses. Also, for $r$ roses, there must be exactly $\frac{50-3r}{2}$ carnations and that number must be an integer, which means $$ 50 - 3 r \cong 0 \pmod{2} \text{,} $$ requiring $r$ is even. Also, every time we include two more roses, we lose three carnations. ... so nine different bouquets, differing in the even number of roses in the interval $[0,16]$ with the remainder of the cost filled out by carnations.
A table of these... \begin{matrix} \text{roses} & \text{carnations} \\ 0 & 25 \\ 2 & 22 \\ 4 & 19 \\ 6 & 16 \\ 8 & 13 \\ 10 & 10 \\ 12 & 7 \\ 14 & 4 \\ 16 & 1 \\ \end{matrix}
A slicker way, using generating functions.
$$ \frac{1}{1-x^3} \cdot \frac{1}{1-x^2} = \cdots + 9 x^{50} + \cdots $$
(How to think about this: $\frac{1}{1-x^3} = 1 + x^3 + x^6 + \cdots$, where the powers track the price that can be obtained by purchasing some number of roses. Similarly, $\frac{1}{1-x^2} = 1 + x^2 + x^4 + \cdots$, where the powers track the price that can be obtained by purchasing some number of carnations. When we multiply these, the powers add, so we get the combined price of some number of roses and some number of carnations. Every combination that has price \$50 contributes one copy of $x^{50}$ to the product. Since there are $9$ copies, there are nine possible bouquets.)
