You should think of this as a two-part problem: first you must pick the $7$ letters to be used, and then you must arrange them. You can’t arrange them until you’ve picked them. Picking them is just picking sets of things: order is irrelevant, so you’re counting combinations. Arranging them, on the other hand, is clearly a matter os specifying an order, so you’re dealing with permutations.
Are there other ways to solve the problem? Yes, but they’re more difficult. You could begin by picking an ordered string of $4$ consonants; this can be done, as you said, in $_7P_4$ ways. You now have a skeleton $s_1C_1s_2C_2s_3C_3s_4C_4s_5$, where $C_1,C_2,C_3$, and $C_4$ are the consonants, and $s_1,s_2,s_3,s_4$, and $s_5$ are the slots into which you can insert vowels. There are now $_5P_3$ ways to select an ordered string $V_1V_2V_3$ of $3$ vowels, and the problem is to count the ways to fit these vowels into the $5$ open slots in the consonant skeleton. Doing that is a matter of selecting a multiset of $3$ not necessarily distinct slots from the $5$ available. This can be done in
$$\binom{5+3-1}3=\binom73={_7C_3}=35$$
ways, so the there are $840\cdot60\cdot35=1,764,000$ such words, exactly the figure obtained by the other computation.
As you noted, "arrange ... in a row" indicates that order does matter, i.e., we are counting permutations. The reasoning you used in parts (a) and (b) is correct. For part (c), one way to count the permutations is
- $2$ ways to choose exactly one of the bride and groom,
- $6$ ways to choose their position,
- $8\cdot7\cdot6\cdot5\cdot4=6720$ ways to choose and arrange the other five people.
Thus, there are $80640=(2\cdot6)\cdot8\cdot7\cdot6\cdot5\cdot4$ permutations in this case.
Best Answer
By definition, $0!=1$. There are a number of different justifications for this. One of them is that this choice makes the formulas you quote give the right answer. One can also argue that there really is exactly one way to choose $0$ objects from $n$: just say no in turn to each of them.
When $k$ and $n$ are non-negative integers, and $k\gt n$, we have two choices for $\binom{n}{k}$. We could say it is undefined. But it is more convenient to adopt the convention that in that situation, $\binom{n}{k}=0$. Again one could give an informal justification, there are $0$ ways to choose $5$ people from $3$. But the real reason for the convention is that it makes some formulas less messy.