Number of ways to arrange people in a row with two restrictions (permutations). The question is as follows:

In a group of 9 people, there are 4 Gujaratis and 5 Marwadis. Of the Gujaratis, 2 are vegetarian and the rest are non-vegetarians. Of the Marwadis, 2 are vegetarian and the rest are non-vegetarians. They have to be seated along a row. No two Marwadis wish to sit together. Furthermore, all vegetarians insist on sitting together. In how many ways can the entire group be seated?
A. $4! \cdot 5! \cdot 4!$
B. $6! \cdot 4! - 2 \cdot 4! \cdot 5!$
C. $6 \cdot 4! \cdot 5!$
D. $6 \cdot 2! \cdot 2! \cdot 2! \cdot 3!$

I have two conflicting approaches to this question. Why is my thinking incorrect?
The letter $G$ represents the Gujaratis and the letter $M$ represents the Marwadis.
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Approach 1
Of the 9 people, 2 Gujaratis and 2 Marwadis are vegetarians. They can be arranged in the following manner: $MGMG$. Therefore, there are $2! \cdot 2!$ ways to arrange the vegetarians.
We can think of these 4 people as one entity. There are 6 places to seat them. Therefore, there are $6!$ ways to seat them.
There are 5 people remaining, 3 Marwadis and 2 Gujaratis. We can notate them in the following way:
$$\_ \ G \ \_ \ G \_$$
There are 3 places for the Marwadis to be seated. Therefore, there are $3! \cdot 2!$ ways to seat the remaining 5 people.
Thus, there are $3! \cdot 2! \cdot 2! \cdot 2! \cdot 6!$ ways to seat all 9 people.
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Approach 2
The group of vegetarians can be grouped as $GMGM$ or $MGMG$. The group of vegetarians can be placed in 6 ways. There are 5 people left who can be arranged in $3! \cdot 2!$ ways. Thus, there are $3! \cdot 2! \cdot 2! \cdot 2! \cdot 6! \cdot 2$ ways to arrange all 9 people.
 A: Your first approach is correct to some extent but is missing something critical.
In this solution, the letter $G$ represents the Gujaratis and the letter $M$ represents the Marwadis.
We will first consider the second restriction - all the vegetarians (2 Gujaratis and 2 Marwadis) insist on sitting next to each other. With the first restriction in mind (no two Marwadis wish to sit next to each other), the 4 people will have to be arranged in one of two ways:

 $$MGMG \text { or } GMGM$$

There are $2!$ ways to seat the Marwadis and $2!$ to seat the remaining two Gujaratis, giving a total of $2! \cdot 2! = 4$ ways to seat all four of the vegetarians in both scenarios.
Let us now consider the group to be $MGMG$ and treat them as an individual entity which we will notate as $V$. Therefore, the vegetarians can be seated in the following ways:

 $$VMGMGM \text { or } MGVMGM \text { or } MGMGVM$$

This gives a total of 3 ways to seat the vegetarians. This leaves us with 3 Marwadis and 2 Gujaratis to be seated. The 3 Marwadis can be seated in $3!$ ways and the 2 Gujaratis can be seated in $2!$ ways.
Therefore, the number of ways to seat all 9 people in this scenario is as follows:

 $$4 \cdot 3 \cdot 3! \cdot 2! = 144$$

However, we must now consider the second option to arrange the vegetarians: $GMGM$. There are still 4 ways to seat all the vegetarians, but they have to be seated differently amongst the entire group.
The vegetarians in this scenario can be seated in the following ways:

 $$MVGMGM \text { or } MGMVGM \text { or } MGMGMV$$

This also gives us a total of 3 ways to seat the vegetarians, and the remaining Marwadis and Gujratis can also be seated in $3! \cdot 2!$ ways.
Therefore, the number of ways to seat all 9 people in this scenario is as follows:

 $$4 \cdot 3 \cdot 3! \cdot 2! = 144$$

Thus, the total number of ways to seat all 9 people is as follows:

 $$144 + 144 = 288 \ (= 6 \cdot 2! \cdot 2! \cdot 2! \cdot 3!)$$

Option D is the correct answer.
The reason why your first approach is incorrect is that you assumed the number of ways to seat the vegetarians was $6!$ when it should have just been 6 (which is the same as the sum of the ways to seat the vegetarians in the $MGMG$ and $GMGM$ scenario). Even if this was a minor typographical error, the reasoning behind why there were 6 ways to seat the vegetarians is incorrect. Since you assumed the vegetarians to be seated as $MGMG$, they could not have been placed in a scenario such as $MGM \_ MG$, where the vegetarians are to be placed in the gap. This would have violated the first restriction.
The reason why your second approach is also incorrect is that you assumed there would be twice as many ways to seat the vegetarians if both the $MGMG$ and $GMGM$ cases were considered, which is not the case. As mentioned in the paragraph above, there are not 6 ways to seat the vegetarians in each scenario; there are only 3 ways. Once again, the same mistake regarding $6!$ is made, when it should have just been $6$.
I know this solution is quite lengthy; I just want to make sure that the reasoning behind this solution is made clear. Nevertheless, I hope you found it helpful.
