In how many ways $50$ different things can be divided among $5$ persons so that, three of them get $12$ things each and two of them get $7$ things each.
The answer is given $$\frac{50!}{(12!)^3\cdot(7!)^2\cdot3!\cdot2!}$$
I cannot understand why.
In my opinion, the answer should be $$\binom{5}{3}\cdot\binom{50}{12}\cdot\binom{2}{2}\cdot\binom{50}{7}\cdot3!\cdot2!\cdot12!\cdot7!$$
I literally cannot comprehend the logic behind the answer given. Is there any easy way to give birth to the answer. I would really like that. Any help is greatly appreciated.
EDIT
The question was just this. According to my understanding, the problem says that suppose there are $5$ persons, where $3$ of them get $12$ things each, meaning all three get $36$ things altogether but in equal amounts, same is the case with remaining $14$ things and $2$ left persons.
$$12+12+12+7+7=50$$
Best Answer
The answer of $\dfrac{50!}{(12!)^3(7!)^23!2!}$ is close but wrong. That is the answer to the related but different question of the number of ways in which we can take 50 distinct items, and break them up into 5 unlabeled piles such that three of the piles specifically have 12 items each and the remaining two piles have 7 items each (this condition was later added in an edit but was missing from the original post).
This can be explained by way of multinomial coefficients and the shepherd's principle. We begin by taking the 50 items and separating them into piles A,B,C,D,E where piles A,B,C are all of size 12 and D,E are each of size 7... and then "forgetting" the labels on the piles to handle overcounting. To "forget" like this, divide by the amount you counted each outcome which is once for each ordering of the labels of A,B,C for the size 12 piles and each ordering of D,E for the size 7 piles.
This answer is incorrect for the current phrasing of the problem in your post for two reasons. First, we are distributing these items to people, not unlabeled piles. People are in combinatorics problems generally always going to be distinguishable, so it absolutely matters to us whether it was Fred who got the first item or if it was Margaret.
Second, nowhere in the original phrasing of the problem does it specify that the two people who aren't picked to receive 12 items are supposed to get 7 items each.
As for your attempt of an answer... I honestly can not understand what you were thinking and it does not resemble anything close to a correct approach. It appears as though you first picked who the three people are who are chosen to get 12 items each... You then for the first person (and apparently only for the first person) decided what twelve items they will get and then you decided what order they appear in that person's pile. Order within piles is generally not considered important, and so you incorrectly applied importance to it here. You then failed to distribute any other items for the other people supposed to get 12. You then went and picked two people from those people remaining (not really necessary since $\binom{2}{2}=1$) and picked what seven items the first of those people gets... but you picked from the overall original list of 50 items without recognizing that 12 of those items are no longer eligible to give since the first person has them already. You then again failed to distribute further and had only dealt with the first person of this category. Lastly, after having picked the people for your categories, you for some reason decided the order in which they are standing which is not relevant to the problem under the most common interpretations.
If we want exactly three people to have 12 items, and the remaining two people to have 7 items.. then the first answer may be modified by assigning piles to people, increasing the count by a factor of $5!$ for a final total of
$$\dfrac{50!\cdot 5!}{(12!)^3(7!)^2\cdot 3!\cdot 2!}$$
If we want at least three people to have 12 items, and the remaining two people to have any number of items including 12 and including zero, I would phrase the answer as $$\binom{5}{3}\binom{50}{12,12,12,14}\left(2^{14}-\binom{14}{12}\right) + \binom{5}{4}\binom{50}{12,12,12,12,2}$$
If we wanted exactly three people to have 12 items, the remaining two people to have any number of items except for 12 but including zero, I would phrase the answer as $$\binom{5}{3}\binom{50}{12,12,12,14}\left(2^{14}-\binom{14}{12}\right)$$