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The problem:

Find the number of ways in which 14 high-school juniors and 10 high-school seniors can be arranged in a line so that no two high-school seniors may occupy consecutive positions

My solution:

I will try find all the permutations where seniors occupy consecutive positions; then, I will subtract the number of permutations found from 24!.

Let us group 10 seniors in groups of 2. So, there will be 5 groups of seniors, each containing two seniors. We will consider one group of seniors as one unit. So, there will be 19 units (14 juniors and 5 groups of 2 seniors). Thus, number of permutations will be 19!.

Now, we will need to find the permutations among the groups of seniors. If we place all the 5 groups of seniors side by side, we will have 10 seniors. So, the number of permutations among seniors will be 10!.

To find the total number of permutations where seniors occupy consecutive positions, we will have to multiply 19! with 10!. To find the answer to our question, we will have to do the following operation,

$$24! - (19! \times 10!)$$

But my answer is incorrect according to my book. My book's answer is $14! \times ^{15}P_{10}$. Why is my answer incorrect?

tryingtobeastoic
  • 3,176
  • Your reasoning is hard to follow. I think you are assuming that the complement consists of arrangements in which $\textit {every}$ senior is next to another senior, but that is not correct. – lulu Jun 18 '20 at 11:57
  • Much easier to arrange the $10$ seniors in a row, creating $11$ blank slots (including those before the first senior and after the last one). We now need to populate those empty slots, the only requirement being that the middle $9$ can not be empty. – lulu Jun 18 '20 at 11:58
  • Your approach seems to involve five pairs of seniors, but even one pair of seniors would break the rule. Instead you could say that every senior must be preceded by a junior (except possibly the first person in line) – Henry Jun 18 '20 at 11:59

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