Basketball Jam

Probability Level 3

How many ways are there to split 10 people into 2 teams so that they can play 5 on 5 basketball?

126 720 120 256 252 1024

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Chung Kevin
Jan 27, 2015

The first guess is that there are ${ 10 \choose 5 } = 252$ ways to split 10 people into a team of 5.

However, notice that whether we form a team of ABCDE, or a team of FGHIJ, we end up with the same 5 on 5 match, namely ABCDE vs FGHIJ. Hence, there are actually only $\frac{252} { 2} = 126$ different way that we can have a 5 on 5 match.

I am tempted to add "The order of the teams doesn't matter". Do you think that would help the problem, or make it worse?

Calvin Lin Staff - 6 years, 4 months ago

Nah, don't add it, then the solver will immediately divide their supposed final answer by $2$ . The solver shouldn't make the assumption that the 2 teams must be distinct: you could have all 5 players change team and it will still be same as Kevin cough has pointed out. There's no specific names given to the two teams, so it goes without saying that the order of the of the team doesn't matter.

Pi Han Goh - 6 years, 2 months ago

Well sorry for being intrusive . I'll just add to your solution to make it complete .

What $\binom{10}{5}$ actually means is that you select 5 things from 10 things (here, things = people) . So when you select a group of 5 people , simultaneously another group of 5 is formed . So if (let's say) you choose ABCDE then simultaneously another group of FGHIJ is formed .

Now included in the $\binom{10}{5}$ selections are the cases when you select ABCDE (the other group formed is FGHIJ) and when you select FGHIJ (the other group formed is ABCDE) , you see you are selecting the same groups of people twice , so we divide by 2 .

A Former Brilliant Member - 6 years, 4 months ago

Laurent Shorts
Apr 10, 2016

Consider the team that person A is playing on. We just have to pick the other $n-1$ people of the team. Answer is then $\left(\begin{array}{c}2n-1\\n-1\end{array}\right)$ .

For $n = 5$ , we obtain ${ 9 \choose 5 } = 252$ .

Vaibhav Gupta
Jan 29, 2015

First we choose 5 people out of 10 people.This is done by 10C5 ways.But when 1 team is formed another team is always created.So we divide by 2.Therefore 252/2=126

Deepak Kumar
Jan 28, 2015

Equivalent to number of ways of forming two groups of 5(i.e group size is 5)out of the 10 players which is given by 10!/((5!)^2)*2!)=126;note that 2! is for two groups of same/identical size.

Basketball Jam

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