There are 4 nickle coins and 4 half nickle coins. How many different options are there for the sum of 5 coins. I have this exercise in combinatorics:

In a drawer there are 4 nickle coins and 4 half nickle coins, bob takes out from the drawer 5 nickles, how many different options are there for the sum of those coins.

In the title I wrote: without systematic elaboration, what I mean by that is:
There are times when I can calculate combinatorics similar questions with multiplications and divisions.
With systematic elaboration(if its the term used in english?) I check all the options "manually":


*

*$1 + 1 + 1 + 1 + 0.5 = 4.5$

*$1 + 1 + 1 + 0.5 + 0.5 = 4$

*$1 + 1 + 0.5 + 0.5 + 0.5 = 3.5$

*$1 + 0.5 + 0.5 + 0.5 + 0.5 = 3$


So there are 4 different sum options.
Most of my combinatorics questions I solve with multiplications and when the order doesn't count I divide, my question is: can this problem be solved with  multiplications and divisions?
 A: All combinatorial problems cannot always be solved by methods other than exhaustive search of all possible states. This problem can't be done with multiplication and division, but it can be solved in the general case without resorting to brute force search of all possible permutations. There are three main cases and I offer explanations of each of them.
By the Pigeonhole principle, at least one nickel and one half nickel coin must be selected. The remaining three coins range from $3\times 0.5$ to $3\times1$. This means there are four possibilities.
If there were $m$ half coins and $n$ full coins, and someone drew out $t$ coins, where $t \gt \max\{m,n\}$, $t \lt m+n+1$ and $t \gt 2$, then by the pigeonhole principle, there must be at least one of each coin. Now consider the remaining $t-2$ spots. Fill them up so as to exhaust any one denomination. Without loss of generality, assume all half nickels are exhausted and the remaining spots are held by full nickels. Now, count the number of full nickels that are NOT selected in this arrangement. Let it be $r$. The number of possible values is now $r+1$ because we could interchange those full nickels with half nickels to get a different value.
Now, if $t < \text{min}\{m,n\}$, then the total number of values possible is simply $t+1$. First, fill all spots with one kind of coin. Then, remove them one by one and replace them with till the spots are filled with the other kind of denomination.
The final case is when $\min\{m,n\} \lt t \lt \max\{m,n\}$. What we do is fill the spots with coins of whichever denomination is maximum. Then, the coins of the other denomination replace the coin one by one. So the number of values is $1 + \min\{m,n\}$.
A: Let $X_1$ be the number of nickels and $X_2$ be the number of half-nickels chosen. We wish to find the number of solutions to $X_1+X_2=5$ subject to the constraints that $0 \leq X_1 \leq 4$ and $0 \leq X_2 \leq 4$ and both are integers.
To start with, assume that the upper bound on the constraints are not present. Imagine 5 stars in a row. Placing one bar somewhere between 2 stars splits the 5 stars into two regions. Let the stars lying on the left of the bar represent $X_1$ and the stars lying to the right represent $X_2$. There are 6 possible places for the bar (if the bar is to the left of the leftmost star, it means $X_1 = 0$), so there are 6 solutions to $X_1+X_2=5$.
Now we consider the upper bounds. We need to subtract the impossible cases from the base count of 6. We do this by the inclusion-exclusion principle. The impossible cases are the ones with $X_1 \geq 5$ OR $X_2 \geq 5$. Consider the $X_1 \geq 5$ case first: the number of cases where $X_1 \geq 5$ and $X_1+X_2=5$ is equal to the number of cases where $X_2 \leq 0$, by combining the inequalities. Clearly there is only one case where this happens: $X_2 = 0$. So we subtract 1 case from the 6. Similarly, there is one impossible case with $X_2 \geq 5$. That leaves 4 possible cases.
The final step of inclusion-exclusion is to add back the cases where $X_1 \geq 5$ AND $X_2 \geq 5$. Fortunately there are none: if both $X_1 \geq 5$ and $X_2 \geq 5$, then combining this with $X_1+X_2=5$ gives $0 \leq -5$, a contradiction. The 4 possible cases remain untouched, so the answer is 4.
This reasoning works with two types of coins, but if a third is introduced such that the same sum can be obtained with two different combinations (e.g. if a three-quarter nickel exists, you can make 1.5 nickels with nickel + half-nickel or two three-quarter nickels) it will not. However it is still usable if you only care about the number of ways you can choose the coins, not their sum.
A: I would just say, any combination you draw must have at least one nickel, and the number of half nickels varies between 1 and 4. The number of half nickels drawn determines the price uniquely, so there are 4 possibilities.
