6 girls and 2 boys are to be seated in a row. How many ways that this can be done if the 2 boys must have exactly 4 girls seated between them.

combinatoricspermutations

The question is the following –

6 girls and 2 boys are to be seated in a row. Find the number of ways that this can be done if the 2 boys must have exactly 4 girls seated between them.

Now, to solve this question, here is my thought process –

  1. There are 3 ways to arrange the boys and girls so that exactly 4 girls are sitting between the boys
    • b g g g g b g g
    • g b g g g g b g
    • g g b g g g g b
  2. Now for each arrangement, the girls can be arranged in 6! ways and the boys can be arranged in 2! ways(i.e. the boys can interchange their positions)
  3. so my final answer comes out to $3 \cdot 6! \cdot 2! = 4320$

However, the book that I follow states that the answer is $288.$ Now I am trying to figure out my mistake

I think that my ways of arranging the boys and girls (the first step) are correct, however, I think I might have made a mistake in arranging the girls. Please help me identify and correct my mistake.

Best Answer

As noted in a comment, your result is the correct answer to the problem as you’ve presented it.

Note that the ratio of the two answers is $15=\binom64$. Thus, $288$ is the number of arrangements if $4$ particular girls need to be sitting between the boys. So it seems that perhaps you missed that part of the question, or the answer given is the answer to a slightly different problem.

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