Sitting Chair problem with 5 boys and 6 girls with restrictions

5 boys and 6 girls are arranged to sit in a row. Only 2 boys are arranged to sit next to each other. How many ways can they sit?

I managed to solve this problem with the following:

$$4*P_6^6*C_4^7*P_5^5$$

$$=12096000$$

So I first sit the 6 girls. Then there would be 7 spaces because of the outermost spaces. Pick 4 out of 7 spaces, and put the 5 boys. Multiply the whole thing by 4, because two of the boys are together there are 4 arrangements, like:

$$12,3,4,5$$ $$1,23,4,5$$ $$1,2,34,5$$ $$1,2,3,45$$

I just felt my method is not very elegant, is there better way to solve this?

Also, more importantly, my friend purposed an alternative approach and it gives wrong answer and I cannot figure out why.

$$P_6^6*P_3^7*C_1^4*P_2^2$$ $$=1209600$$

The result is exactly ten times less than the correct answer.

So the thought process is to firstly sit the six girls. Then out of the seven spaces, put down three individual boys. There would be 4 spaces left, so pick one, and put in the two boys and these two boys can swap.

Why doesn't this work?

Solution 1:

The wording of the question is confusing to me. The question says Only 2 boys are allowed to sit next to each other. It does not say that two boys have to necessarily sit next to each other.

Based on your interpretation, your first approach is a good way to solve the problem.

$ \displaystyle 6! \cdot {7 \choose 4} {4 \choose 1}\cdot 5!$

You choose $4$ spaces out of $7$ between girls to seat the boys and from those $4$ spaces, choose $1$ where you would places $2$ boys together.

In your second approach, you also need to choose $3$ boys who would be sitting individually and note that ${5 \choose 3} = 10$, the missing factor in your answer, besides the typo I called out in comments. So the answer is,

$ \displaystyle 6! \cdot {5 \choose 3} \cdot {7 \choose 3} \cdot 3! \cdot {4 \choose 1} \cdot 2! = 12096000$