Partial answer
As said in comments the answer to $1)$ is $\binom{7}{1}\cdot \binom{5}{1} \cdot \binom{2}{1}.$
The answer to $2)$ is $\binom{7}{2}\cdot \binom{5}{2} \cdot \binom{2}{2}$ (you choose two vegetables and you have $\binom{7}{2}$ possible ways to do it, and similar for meat and cheese. Remember that you have to multiply).
The answer to $5)$ is
$$\binom{7}{2}+\binom{7}{3}+\binom{7}{4}+\binom{7}{5}+\binom{7}{6}+\binom{7}{7}.$$ If you choose two vegetables you have $\binom{7}{2}$ ways to do it, If you choose three vegetables you have $\binom{7}{3}$ ways to do it, and so on. Use that
$$\binom{7}{2}+\binom{7}{3}+\binom{7}{4}+\binom{7}{5}+\binom{7}{6}+\binom{7}{7}=2^7-\binom{7}{0}-\binom{7}{1}$$ to get its value.
Could you get the solution to the other questions?
Rephrasing the question into a more common form, you are asking for the number of integral solutions to the following system:
$$\begin{cases} x_1+x_2+x_3 = 6\\ 0\leq x_1\leq 3 \\ 0\leq x_2\leq 4 \\ 0\leq x_3\leq 5\end{cases}\\\text{where each of}~~x_1,x_2,x_3\in\mathbb{Z}$$
You correctly noted above that if there was not a maximum limit above that the solution would be $\binom{6+3-1}{6}$, as per the stars-and-bars method.
We will couple this knowledge with the principle of inclusion-exclusion. The combinations we are interested in will be the set of all possible combinations without restriction minus those combinations which violate one or more of the restrictions.
Let $A_1$ be the set of all combinations such that you violate the upper bound on $x_1$ (i.e. you have strictly more than 3 chicken dishes). How many combinations exist in $A_1$ that we will need to subtract from the total?
Well, since the upperbound condition is violated, that means $4\leq x_1$. By a change of variable $y_1=x_1-4$, we arrive at the system:
$$\begin{cases} y_1+x_2+x_3 = 2\\ 0\leq y_1 \\ 0\leq x_2 \\ 0\leq x_3\end{cases}\\\text{where each of}~~y_1,x_2,x_3\in\mathbb{Z}$$
The number of which then is $\binom{2+3-1}{2}$.
Continue calculating $A_2$ and $A_3$ which will represent having violated $x_2$'s upper bound and $x_3$'s upper bound respectively (i.e. having sold too many beef and lamb dishes).
In general though, it is possible for these to intersect (that you could have simultaneously sold too many chicken and simultaneously sold too many lamb dishes). You will need then to calculate $A_{1,2}, A_{1,3}, A_{2,3}$ and possibly even $A_{1,2,3}$ denoting having violated each of the corresponding conditions simultaneously.
Letting $B$ denote no violated conditions, and $S$ denoting no conditions present in the first place, by principle of inclusion-exclusion the answer will be:
$|B| = |S| - |A_1| - |A_2| - |A_3| + |A_{1,2}| + |A_{1,3}| + |A_{2,3}| - |A_{1,2,3}|$
Which in this case is:
$=\binom{6+3-1}{6} - \binom{2+3-1}{2} - \binom{1+3-1}{1} - \binom{0+3-1}{0} + 0 +0 + 0 - 0$
(noting that it is impossible in this case to simultaneously sell too many of two types of dish at the same time)
To generalize this problem to the larger case, let $A_{i}$ denote violating the $i^{th}$ condition (and possibly more), $A_{i,j}$ denote violating the $i^{th}$ and $j^{th}$ conditions, ... $A_{i,j,\dots,p}$ denote violating each of the conditions $i$ through $p$, and even more generally for an index set $\Delta \subset \{1,2,\dots,k\}$ you will have $A_\Delta$ violating all of the conditions on $x_i$ for each $i\in\Delta$:
$$|B| = |S| + \sum\limits_{i=1}^k\left((-1)^i\sum\limits_{|\Delta|=i}|A_{\Delta}|\right)$$
And $|A_\Delta| = \binom{n +k - 1- \sum\limits_{i\in\Delta}(r_i+1)}{n}$, where $r_i$ is the upper bound for $x_i$
Best Answer
There are $10$ possible toppings on a pizza, and each of these toppings individually has $2$ different possibilities: on the pizza, or off the pizza. Hence the total number of possible pizzas is $2^{10}$.
I like to visualize it as $10$ "slots", one for each topping, and the number of options each individual topping can have - i.e. $2$, because each individual topping is a simple "yes" or "no" - goes into each slot. Then all $10$ of those $2$'s get multiplied to arrive at the final answer of $2^{10}$.
Notice this also counts the pizza with no toppings at all: it is simply the option where you have chosen "no" for every one of the $10$ options.
You would use factorials if you wanted to count permutations of toppings - but, in this case, the order you put the toppings on the pizza does not matter. You get the same pizza (presumably) whether you put the pepperoni on first or the mushrooms on first.