1
$\begingroup$

I'm trying to improve my knowledge of statistics and develop my intuition for solving statistical problems. While doing so I've worked on the following exercise:

There are 20 players in a checkers tournament. How many pairings of players could there be, if the colour of the player - red or black - matters?

I've solved this problem in two ways and arrived at the same result. The result is quite large, which leaves me to ask - is my intuition correct, or an I making a common mistake?

Approach 1: There must be 10 red players in this tournament, each player can only play one game at a time and each one could sit at any chess set. Therefore, I must choose 10 players from 20 without replacement: $ \frac{20!}{10!} $. The 10 remaining players must all be black and we don't need to account for these because, in selecting who is red and who sits where, we have accounted for all possible pairings between red and black players. Therefore there are 670,442,572,800 possible match-ups.

Approach 2: Any 10 of the 20 people must be red, There are $ {20}\choose{10} $ subsets of the players with size 10, which is representative of all of the subsets of the players that could have chosen to be red. All of the 10 remaining players must choose who to play, and two black players can't choose the same red player to play. So, there are $ 10! $ possible ways that the black players could choose red players. Applying the multiplication rule we arrive at 670,442,572,800 possible match-ups.

The key difference between these solutions is where I introduce the notion of ordering. Is the answer I've arrived at correct and is the reasoning I've applied to solve this problem sound?

$\endgroup$
2
  • $\begingroup$ PS: this isn't a homework question; I'm working on this solely to improve my understanding of mathematics. $\endgroup$
    – Liam M
    Mar 13, 2016 at 7:13
  • 3
    $\begingroup$ The expression $\frac{20!}{10!}$ and the equivalent $\binom{20}{10}10!$ are right. $\endgroup$ Mar 13, 2016 at 7:21

1 Answer 1

2
$\begingroup$

Okay, let's try another approach.

Approach 3: We let the players choose the seat they want to sit in. They can do it in $20!$ ways. Now, we can apply a color, say red to the players that sit at the left, and black to the ones on the right. We need not consider the opposite coloring as we've already considered all the possible permutations. So, the number of such pairs remains unchanged. Now, as the tables are identical, we need to divide the count by $10!$ because the same pairing can would occur in $10!$ cases. So, this gives me $\frac{20!}{10!}= 670,442,572,800$. It's the same as you got. So, I think you were right.

$\endgroup$
2
  • 1
    $\begingroup$ Thanks for the extra approach! $\endgroup$
    – Liam M
    Mar 13, 2016 at 8:50
  • $\begingroup$ @LiamM You're welcome! $\endgroup$ Mar 13, 2016 at 10:28

You must log in to answer this question.

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