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.
You can seat Mrs Jones in any of $8$ places. Once she’s seated, there are $3$ places that are not available to Mr Jones: the place where she’s seated, and the two adjacent places. Thus, there are $5$ choices for his seat. That leaves $6$ people who can fill the remaining $6$ seats in any of $6!$ possible orders, for a total of $8\cdot5\cdot6!=28,800$ possible arrangements if the seats are the table are individually identified.
If we care only about who sits next to whom, then we don’t care which of the $8$ seats Mrs Jones takes: all we care about is where the others sit in relation to her. That reduces the number of choices to $5\cdot6!=3600$ and must therefore be the intended interpretation.
Best Answer
Take the word SIGNATURE, and find permutations if
(a): no two vowels are together (used gap method)
(b): The letters S,G,N are together (used string method)
(a) There are $4$ vowels which can be placed in the gaps between consonants in $\binom64$ ways, and the vowels and consonants can be permuted separately.
$- S - G - N - T - R - \;\; \binom64 \times 4! \times 5!$
(b) Clump $SGN$ into a string $\boxed {SGN}$ so there are $1 + 6 =7$ "objects"
The clumped string can be permuted in $3!$ ways, and the $7$ objects in $7!$ ways, thus $7!*3!$ ways
There can be many variations of such problems, and all letters may not be distinct as they are here, but the basic principles of the two methods will remain the same.