5
$\begingroup$

There are $2n$ people, each with a unique height. I am to place them in two rows of $n$ people in each row such that the person in the front row is shorter than the one directly behind. How many ways can I do this?

I am assuming that order of how the people are placed matters (for example, row one: 1,2,3 is different from 2,3,1). I'm thinking about this problem as the number of ways to place the $2n$ people in pairs, such that the order of picking pairs matters. Thus, my answer should be $${2n \choose 2} \cdot {2n-2 \choose 2}\cdot \cdots\cdot {2 \choose 2}.$$

Is this method of thinking correct? I've tried to see if this works for $n=1,2$ and it seems to hold up.

$\endgroup$
2
  • 3
    $\begingroup$ Yes, this can be written $\frac{(2n)!}{2^n}$. $\endgroup$ Commented Oct 16, 2017 at 23:47
  • $\begingroup$ Are the rows distinguishable? $\endgroup$
    – Paul
    Commented Oct 16, 2017 at 23:49

1 Answer 1

1
$\begingroup$

Yes, it is correct.

First, let the lines (rows) be like that:

$a_1\ a_2\ a_3\ a_4\ \cdots\ a_n$

$b_1\ b_2\ b_3\ b_4\ \cdots\ b_n$

Such that $a_k$ is in the front row and $b_k$ is the person behind.

Thinking Process:

First you choose $a_1$ and $b_1$ in $2n$ people. There will be ${2n \choose 2}$ choices.

Then, there is $2n - 2$ people left, and you need to choose $a_2$ and $b_2$, hence ${2n-2 \choose 2}$, and so on...

Then by the product rule, this is the answer: $${2n \choose 2} \cdot {2n-2 \choose 2}\cdot \cdots\cdot {2 \choose 2}$$ or $$\prod_{k=1}^{n} {2k \choose 2}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .