Filling a bag with fruits of four types, with constraints Here's a diabolical math problem that I found. 
In how many ways can we fill a bag with n fruits subject to the following constraints?
• The number of apples must be even.
• The number of bananas must be a multiple of 5.
• There can be at most four oranges.
• There can be at most one pear.
This problem is purely for recreational purposes, but I prefer to see a solution that involves "pure" counting, and not generating functions.
 A: Here's a tedious 'solution by counting'.
Let $S_{op}(n)$ denote the number of solutions to the original problem. Let $S_o(n)$ denote the solution to the problem, except no pears are included in the bag. Let $S(n)$ denote the solution to the problem with no pears or oranges in the bag, which is to say, $S(n)$ is the number of nonnegative integer solutions $(a,b)$ to the equation 
$$2 a + 5 b = n $$
Notice that $$S_{op}(n) = S_o(n) + S_{o}(n-1)$$, as we count in the first summand how many ways there are to bag exactly $n$ fruits with no pears, and in the second summand we count how many ways there are to bag $n-1$ fruits with no pears, it being implicit that we intend to sneak a pear in afterwards to make a full $n$.
In a similar vein,
$$
S_o(n) = S(n) + S(n-1) + \cdots + S(n-4)
$$
In doing so, we have shown
$$
S_{op}(n) = S(n) + 2 \sum_{k = 1}^4 S(n- k) + S(n-5)
$$
where, to make the formula true in all cases (for small $n$), we adopt the convention $S(n) = 0$ for $n = 1,0,-1,\cdots$.
We now count $S(n)$; I derived that we have
$$
S(n) = 
\begin{cases}
\left\lfloor \frac{n}{5} \right\rfloor & n \text{ odd} \\
\left\lfloor \frac{n}{5} \right\rfloor + 1 & n \text{ even}
\end{cases}
$$
Putting all this together gives a solution to the original problem $S_{op}(n)$.
A: We want to count the number of solutions in integers of 
$2 x_1 + 5 x_2 + x_3 + x_4 = n$
subject to $x_i \ge 0$ for all $i$, $x_3 \le 4$ and $ x_4 \le 1$.
As suggested by @oldrinb, this problem has an elegant solution via generating functions.  Let $a_n$ be the number of solutions, and define $f(x) = \sum_{n=0}^{\infty} a_n x^n$.  Then "clearly" we have
$f(x) = (1+x^2+x^4+\dots) (1+x^5+x^{10}+\dots) (1+x+x^2+x^3+x^4) (1+x)$
$= \frac{1}{1-x^2} \cdot \frac{1}{1-x^5} \cdot \frac{1-x^5}{1-x} \cdot (1+x)$
$= (1-x)^{-2}$
$= \sum_{n=0}^{\infty} (n+1) x^n$
So $a_n = n+1$.
Given the simplicity of the result and the OP's request for a non-GF solution, we seek a combinatorial proof that the number of solutions is $n+1$.  To that end, suppose $n \ge 0$ is given.  Consider one of the $n+1$ pairs $(i, n-i)$, $i = 0, 1, 2, \dots ,n$.  There are unique non-negative integers $x_1$ and $x_4$, $x_4 \le 1$, such that $2 x_1 + x_4 = i$.  And there are also unique non-negative integers $x_2$ and $x_3$, $x_3 \le 4$, such that $5 x_2 + x_3 = n-i$.  Then $2 x_1 + 5 x_2 + x_3 + x_4 = i + (n-i)= n$. Given a solution of the equation, we can also easily generate the pair $(i, n-i)$; so we have a bijection between the solutions of the equation and the $n+1$ pairs $(i, n-i)$.
