$10$ students ordering $10$ different combinations of $5$ dishes so that each dish appears in at least $1$ combination 
In how many ways can $10$ students in the queue for sandwiches order $10$ different combinations of $5$ side dishes so that every dish is in at least $1$ combination? Every combination can consist of $0$ to $5$ (included) dishes.

I tried to find some more general set of examples, so to speak. I viewed different combinations as different sets with at most $5$ different elements/dishes. Those sets are different if they either have different cardinality or equal cardinality, but different elements.

*

*$\binom{5}5=1\implies$ at most $1$ student could order all $5$ dishes.

*$\binom{5}4=5\implies$ at most $5$ students can order $4$ dishes.

*$\binom{5}3=10\implies$ all $10$ students can order $3$ dishes.

*$\binom{5}2=10\implies$ all $10$ students can order $2$ dishes.

*$\binom{5}1=5\implies$ at most $5$ students can order only $1$ dish.

*$\binom{5}0=1\implies$ at most $1$ student can order nothing

Let $x_i\ge 0$ be the number of students who ordered $i$ dishes, $i\in\{0,1,\ldots,5\}.$  Then, we have $$x_0+x_1+x_2+x_3+x_4+x_5=10$$ and the following restrictions $$x_0\le 1, x_1\le5,x_2\le10,x_3\le10,x_4\le 5,x_5\le1.$$
Since $x_i\le\text{something},$ I couldn't use the substitution $t_0=x_0-\text{something},$ so, I instead tried to use the method of complements, hence, for example, instead of $x_3\le 10$ I would take $x_3\ge 11,$ analogously for all $i$ sum the solutions an substract them from the solutions of $x_0+x_1+x_2+x_3+x_4+x_5=10,$ with no restrictions other than $x_i\ge 0$ and make use of the result:

There are $\binom{n+k-1}k\quad n$ tuples of nonnegative integers satisfying $x_i\ge0\forall i\in\{1,\ldots,n\}.$

However, something went wrong here and I have problems with every dish being in at least $1$ combination and I don't know how to proceed. How should I deal with this problem?

To be honest , as i wrote in comment section , i could not clearly understand the question.Maybe ,it is because of my english. Anyway , it seems that OP understood their question and produced a solution way ,but he suffer from calculating the equation of $$x_0 +x_1 +x_2 +x_3 +x_4 +x_5 =10$$ where $x_i \geq 0 , x_0 \leq 1 , x_1 \leq 5  ,x_2 \leq 10 , x_3 \leq 10 ,x_4 \leq 5 , x_5 \leq 1$
He tried "complemet " method , but he cannot achieve it , so i recommend him to use generating functions such that

*

*Generating function of $x_0$ is equal to $$x^0 +x^1 = \frac{x^2-1}{x-1}$$


*Generating function of $x_1$ is equal to $$x^0 +x^1 +x^2 +x^3 +x^4 +x^5= \frac{x^6-1}{x-1}$$


*Generating function of $x_2$ is equal to $$x^0 +x^1+..+x^9 +x^{10} = \frac{x^{11}-1}{x-1}$$


*Generating function of $x_3$ is equal to $$x^0 +x^1+..+x^9 +x^{10} = \frac{x^{11}-1}{x-1}$$


*Generating function of $x_4$ is equal to $$x^0 +x^1 +x^2 +x^3 +x^4 +x^5= \frac{x^6-1}{x-1}$$


*Generating function of $x_5$ is equal to $$x^0 +x^1 = \frac{x^2-1}{x-1}$$
Now , find the coefficient of $x^{10}$ in the expansion of $$\frac{x^2-1}{x-1} \times \frac{x^6-1}{x-1} \times \frac{x^{11}-1}{x-1} \times \frac{x^{11}-1}{x-1} \times \frac{x^6-1}{x-1} \times \frac{x^2-1}{x-1}  $$
CALCULATION
So , answer is $721$
$\mathbf{\text{EDITION=}}$ I now understand that question ask that how many ways are there to distribute $5$ different dishes to $10$ students such that a student can take at least $0$ and at most $5$ dishes. It is classical distributing distinct objects into distinct boxes problem. To solve it , the most suitable method is exponential generating functions.
Because of the restriction , we know that the exponential generating function of each student is  equal to $$\bigg(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\bigg)$$
Now , because of there are $10$ students , find the expansion of $$\bigg(1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!}\bigg)^{10}$$
After you find the expansion , find the coefficient of $\frac{x^5}{5!}$ or find the coefficient of $x^5$ and multiply it by $5!$
LOOK AT FOR COMPUTATION
Then , the answer is $$\frac{2500}{3} \times 5! = 100,000$$
