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.
a) this one depends on the question at hand.
Say you have 3 people (A, B, C) and you want to put them in a row. Finding all the arrangements would include
A B C
A C B
B C A
B A C
C A B
C B A
3P3 = 3! = 6 arrangements
but the same three people were in there so if order didn't matter, this would only count as 1 arrangement since in each case there was 1A, 1B, 1C.
that's how a combination is different
$3\choose3$ = 1 arrangement
b) You're right that's clever. Locks don't have permutations or combinations actually
you can't say a lock has 10P3 = 10 x 9 x 8
A lock has different numbers for each position
so n(S) = 10x10x10
c) The combination is just the number of choices provided we are not ordering the items but choosing a certain amount of them and grouping them together. A permutation is all the ways of arranging all the combinations into specific orders like in a)
A question on combinations.
If there are 30 people in a class and you need to pick 2 people to clean up at the end of the day. How many ways can you choose those people. This is one where order doesn't matter because picking Eddie and Fred is the same as Fred and Eddie
n(S) = $30\choose2$ = 435 (not 30P2 = 870. Note that 870 is double 435 so see how in this case you would have had 870 full arrangements but only half of them are valid choices. Why is that? Obviously because for every Fred and Eddie there was a Fred Eddie)
If you are choosing 3 people it gets worse because
30P3 = 24 360 but $30\choose3$ is 4 060. Clearly you cannot just divide the permutation by 3 to get 4060. You need to divide it by 3! because there will be
Fred, Eddie, Freddie.
Fred, Freddie, Eddie.
Eddie, Fred, Freddie.
Eddie, Freddie, Fred.
Freddie, Fred, Eddie.
Freddie, Eddie, Fred.
$A\chooseB$ = $A\PermuteB$ / B!
Best Answer
Here, as an example, $$P(5, 3) = \frac{5!}{2!}$$ means that out of 5 possible objects, how many ways there are to choose 3 objects where order matters. On the other hand, $$C(5, 3) = \frac{5!}{3!\cdot 2!}$$ means the same thing except order doesn't matter.
In this case, note that the textbook says "where order matters." Hence, $P(N, R)$ is number of ways to choose R objects where order matters and $C(N, R)$ is the number of ways to choose an object where order doesn't matter.
Does this help?