Combinatorics exercise about sets of forks, knives, spoons. Problem statement:
At various yard sales, a woman has acquired five forks, of which no two are alike. The
same applies to her four knives and seven spoons. In how many different ways can
three place settings be chosen if each place setting consists of exactly one fork, one
knife, and one spoon? Assume that the arrangement of the place settings on the table is
unimportant.
So far I figured the following thing out:
Since we have to serve 3 sets, having the sets already, we would have $3!=6$ different ways of arranging the already created sets.
What would be the next thing to calculate to count how many times ONE set could be arranged?
The part I am struggling with is that each fork (and knive, spoon respectively) is distinct from one another, so I can't just use this theorem:

It would be nice if somebody would only give me a hint so I can still figure it myself (Dont just handwaive). Thank you!
 A: You asked for a hint:
Hint: Pick a fork. Pick a knife to go with it and a spoon to go with it; repeat; repeat again. Then remember that (according to the problem statement) the order in which sets are chosen doesn't matter, so this process overcounts, and that must be corrected for.
A: Well, for the first place setting to choose the fork we have $\binom{5}{1}=5$ choices. Now given that choice, we have $\binom{4}{1}=4$ choices for the knife and finally given that fork and knife, we have $\binom{7}{1}=7$ choices for the spoon. Hence there are $$5\cdot 4\cdot 7=140$$ different ways to choose the first placemat. Once this placemat is chosen, we certainly can not choose that same fork, knife, and spoon again can we? So what must be modified?
A: Alternative approach:
My approach is linear, and is the approach that I recommend for students new to this area of Math.
The distribution is a two step process:

*

*Step 1:  You have to select the $3$ knives, forks, and spoons.


*Step 2: Designating the knives as knife-1, knife-2, knife-3, you have to associate the forks and spoons with each of the $3$ knives.
For Step 1:
The enumeration is
$$\binom{5}{3} \times \binom{4}{3} \times \binom{7}{3} = 1400.\tag1 $$
For Step 2:
The distribution of the forks to knife-1, knife-2, knife-3 can be permuted in $(3!)$ ways.  The same for the spoons.  Therefore, the enumeration here is
$$(3!)^2 = 36.\tag2 $$
Putting (1) above and (2) above together, the overall enumeration, based on the assumption that it does not matter which place setting is located at any specific table location is:
$$1400 \times 36 = 50400.$$
