Probability that a group of 7 randomly choosen pizzas includes 4 specific pizzas when there are 4 kinds of pizzas to choose from. This Problem is from "Bayesian Reasoning and Machine Learning" Exercise 1.17

Seven friends decide to order pizzas by telephone from Pizza4U based
on a ﬂyer pushed through their letterbox. Pizza4U has only 4 kinds of
pizza, and each person chooses a pizza independently. Bob phones
Pizza4U and places the combined pizza order, simply stating how many
pizzas of each kind are required. Unfortunately, the precise order is
lost, so the chef makes seven randomly chosen pizzas and then passes
them to the delivery boy.>

*

*How many different combined orders are possible?


*What is the probability that the delivery boy has the right order?

If I'm not mistaken the number of different orders is $4^4 = 64$ but I don't know how to do 2. Any help would be greatly appreciated.
 A: For question $1$, I assume you meant $4^7$ rather than $4^4$?  (for there are $7$ people that can each order one of $4$ pizzas).
Anyway, that doesn't work as an answer to question $1$, because it would be $4^7$ only if the order of the people matters, but it doesn't.  For example, if Bob, Alice, Carrie, and Dave, Edgar, Felice, and Greg order (in the order as listed) pizzas $A,B,C,B,C,C,D$ respectively, then that it is in effect the same pizza order as when they would have ordered $B,C,B,A,C,D,C$:  it is still $1$ pizza $A$, $2$ pizza's $B$, $3$  pizza's $C$, and 1 pizza $D$.
For part $2$, you need to consider the chance of each of the possible orders being made and ordered.  Note that we are assuming that the 7 friends chose their pizzas randomly as well (not very realistic, but hey ...).  Anyway, as an example:
The pizza order $7$ pizza's $A$ has a chance of $1$ in $4^7$ of being ordered and has the same chance of being made. So, a match for that particular order is $1$ in $4^{14}$
But the pizza order I gave as an example earlier, $1$ pizza $A$, $2$ pizza's $B$, $3$  pizza's $C$, and 1 pizza $D$ has a much larger chance of being ordered and made: 7 different people could have ordered pizza $A$, and from the remaining $6$, there are ${6 \choose 2} = 15$ pairs of people that can order pizza $B$, etc.  So the chance of this order being ordered and made is:
$1$ in $( {7 \choose 1}\cdot {6 \choose 2} \cdot  {4 \choose 3} \cdot {1 \choose 1})^2$
So this is what you need to do for all possible orders.  As @S.Ong said, it's tedious.  (though mathematicians more clever than I am will probably know some better method that is far quicker than this ...)
A: *

*Note that the order of pizzas does not matter; as such, you overcount.

*If the pizzas are independently sequentially chosen, the probability of success depends upon the order itself (consider the cases where all 7 are the same pizza, or when one is different from the other 6). This will likely require some messy case work. If the random selection is from the set of all distinct combinations of pizzas, then your probability space is uniform, so you can use your answer from part 1) here.
A: *

*A nice way to show this is to consider using ‘dividers’ that separate the $7$ friends into their pizza choices. For example $oo|ooo|o|o$
would say that two have chosen pizza A, three pizza B, one pizza C and one pizza D. We are interested in the number of such
partitions. There are then $7 + 3 = 10$ positions, each position containing either a friend $o$ or a divider $|$. We can place the dividers
in $10 × 9 × 8 = 720$ positions. Since the dividers have themselves $3 × 2 = 6$ arrangements, the number of partitions is $720/6 = 120$.


*Note that this is not $1/120$ – this would be the probability of the chef uniformly choosing one of the $120$ partitions. However, he
chooses a pizza at random, so that some partitions are more probable than others. For example, it is less likely that all friends chose
pizza A, than two friends choosing pizza A, two friends pizza B, two friends pizza C and one friend pizza $D$. One needs therefore
to compute these probabilities, $p_i, i = 1,..., 120$. Since the chef and friends chose independently, the probability of agreement is
given by $\sum_i^{120} p^2_i ≈ 0.0184.$
Source of this answer is this solution manual
