Which approach is correct and why and why other is wrong

Question

The question is as follows

For a game in which every pair has to play with every other pair than find total number of games played if total players is 8

Method 1 - divide 8 into 4 groups of 2 and than selecting 2 grips out of 4 using ${(8!)÷(2!×2!×2!×2!×4!)}$ ×$4\choose2$=630
Method 2 - select 4 players out of 8 and than divide in 2 groups of 2 using $8\choose4$×${(4!)÷(2!×2!×2!)}$=210

I think in 1st case repetitions occur but can't see how or is there anything else wrong

In the first case you are also splitting the 4 non-playing people into two groups of two. This can be done in 3 ways, so you are counting every match three times. — Jaap Scherphuis, Nov 06 '19 at 14:23

score 1 · Accepted Answer · answered Nov 06 '19 at 14:26

An easier approach is to say there are ${8 \choose 2}=28$ ways to choose the first pair for the game, ${6 \choose 2}=15$ ways to choose the second pair, and divide by the $2$ orders we can get the same pairs. That gives $210$. There is no need to refer to the other players. This is equivalent to your method $2$.

For method 1 you divide into $(AB),(CD),(EF)(GH)$ and play $(AB),(CD)$, then also divide into $(AB),(CD),(EH)(FG)$ and play $(AB),(CD)$. You count each pairing three times for the different ways to divide the other players.

Which approach is correct and why and why other is wrong

1 Answers1