Our basketball team has 10 players, including Steve and Danny. We need to divide into two teams of 5 for an intra-squad scrimmage. In how many ways can we do this without restrictions?
I'm not sure I know how to do this. It isn't 10 choose 5.
[Math] Our basketball team has 10 players. We need to divide into two teams of 5 for an intra-squad scrimmage. In how many ways can we do this
combinatorics
Related Solutions
a) Pick who the referee is in $11$ ways. Then among the ten people leftover, one will be the youngest. Pick who the other four members of that player's team is in $\binom{9}{4}$ ways. We get a total of $11\binom{9}{4}=1386$ ways.
b) Pick who the referee is in $11$ ways. Simultaneously pick the two captains in $\binom{10}{2}$ ways. Pick who the other four members of the youngest captain's team is in $\binom{8}{4}$ ways. We get a total of $11\binom{10}{2}\binom{8}{4}=34650$ ways.
Alternate approach:
a) Arrange five a's, five b's, and a c in a line in $\frac{11!}{5!5!1!}$ ways. Have the players go to either team $A$, team $B$, or referee position according to the arrangement of the a's, b's, and c's compared to their names. Now, "forget" which team was actually team $A$ and which team was team $B$ by dividing by two. This gives us $\frac{11!}{5!5!1!}\cdot\frac{1}{2}=1386$ ways.
b) Arrange four a's, four b's, one A, one B, and one C in a line in $\frac{11!}{4!4!1!1!1!}$ ways. The a's will correspond to which players are normal members of team $A$, the A will correspond to the player who is captain for team $A$ etc... Finally, "forget" which team was actually team $A$ and which was team $B$ by dividing by two. This gives us $\frac{11!}{4!4!1!1!1!}\cdot\frac{1}{2}=34650$ ways.
Similarly, one could instead just take the answer to part (a) and then pick who from the youngest person's team is captain and pick who the captain is for the other team, making it so there is $5\cdot5\cdot1386=34650$ ways.
In how many ways can nine students be divided into teams of $4$, $3$, and $2$ people?
The teams are distinguished by their sizes. Choosing who is on each team completely determines the teams.
There are $\binom{9}{4}$ ways to select four of the nine students to be on the team of four students, $\binom{5}{3}$ to select three of the five remaining students to be on the team with three students, and one way to form a team of two from the remaining two students. Hence, there are $$\binom{9}{4}\binom{5}{3} = \frac{9!}{4!5!} \cdot \frac{5!}{3!2!} = \frac{9!}{4!3!2!}$$ ways to divide the nine students into three unlabeled teams.
If we had instead chosen the team of two, then the team of three from the remaining seven students, and then placed the remaining four students on the team of four, we could select the teams in $$\binom{9}{2}\binom{7}{3} = \frac{9!}{2!7!} \cdot \frac{7!}{3!4!} = \frac{9!}{2!3!4!}$$ ways, in agreement with above.
Notice that labeling the team with four students team A, the team with three students team B, and the team with two students team C would not change our answer.
More care would be required if two or more of the groups had the same size.
Suppose our students are Amanda, Brenda, Claire, Dennis, Edward, Fiona, Gloria, Henry, and Ivan.
In how many ways can nine students be divided into three unlabeled teams of three people?
If we divide the nine students into teams of three, then the $3! = 6$ divisions \begin{align*} & \{Amanda, Brenda, Claire\}, \{Dennis, Edward, Fiona\}, \{Gloria, Henry, Ivan\}\\ & \{Amanda, Brenda, Claire\}, \{Gloria, Henry, Ivan\}, \{Dennis, Edward, Fiona\}\\ & \{Dennis, Edward, Fiona\}, \{Amanda, Brenda, Claire\}, \{Gloria, Henry, Ivan\}\\ & \{Dennis, Edward, Fiona\}, \{Gloria, Henry, Ivan\}, \{Amanda, Brenda, Claire\}\\ & \{Gloria, Henry, Ivan\}, \{Amanda, Brenda, Claire\}, \{Dennis, Edward, Fiona\}\\ & \{Gloria, Henry, Ivan\}, \{Dennis, Edward, Fiona\}, \{Amanda, Brenda, Claire\} \end{align*} are all equivalent since they result in the same three teams. Therefore, the number of ways of dividing the class into three unlabeled teams of three is $$\frac{1}{3!}\binom{9}{3}\binom{6}{3} = \frac{1}{3!} \cdot \frac{9!}{3!3!3!}$$ We divide by $3!$ to account for the $3!$ orders in which we could select the same three teams of three.
In how many ways can the nine students be divided into three unlabeled teams of sizes $2$, $2$, and $5$?
Similarly, if the teams are not labeled and we divide the class into two teams of two and one team of five, the two divisions \begin{align*} \{Amanda, Brenda\}, \{Claire, Dennis\}, \{Edward, Fiona, George, Henry, Ivan\}\\ \{Claire, Dennis\}, \{Amanda, Brenda\}, \{Edward, Fiona, George, Henry, Ivan\} \end{align*} are equivalent since they result in the same three teams. Hence, the number of ways of dividing the nine students into two teams of two and one team of five if the teams are unlabeled is $$\frac{1}{2!}\binom{9}{2}\binom{7}{2} = \frac{1}{2!} \cdot \frac{9!}{2!2!5!}$$ We divide by $2!$ to account for the $2!$ orders in which we could pick the same teams of size two.
If we had instead picked the team of five first, we would be left with four people. You might think that the two teams of two could be picked in $\binom{4}{2}$ ways, but this counts every team twice, once when we choose a team and once when we choose its complement. Alternatively, notice that if our team of five consists of Edward, Fiona, Gloria, Henry, and Ivan, the two teams of two are distinguished by who is paired with Amanda. There are three ways to do this: \begin{align*} \{Amanda, Brenda\}, \{Claire, Dennis\}\\ \{Amanda, Claire\}, \{Brenda, Dennis\}\\ \{Amanda, Dennis\}, \{Brenda, Claire\} \end{align*} Hence, the number of divisions of nine students into two teams of two and one team of five is $$\binom{9}{5} \cdot 3 = \binom{9}{5} \cdot \frac{1}{2}\binom{4}{2} = \frac{1}{2}\binom{9}{5}\binom{4}{2} = \frac{1}{2!} \cdot \frac{9!}{2!2!5!}$$
Notice that the team of five is distinguished by its size, while the two teams of two are not. The teams of two can only be distinguished by who is on which team.
To summarize, teams of different sizes are distinguished by their sizes, so the order in which they are selected does not matter. If we have unlabeled teams of the same size, we have to divide by the number of orders in which we could pick the same teams.
Best Answer
Choosing players
Team 1: ABCDE
Team 2: FGHIJ
leads to the exact same matchup as
Team 1: FGHIJ
Team 2: ABCDE.
So the answer is $\frac{1}{2} \binom{10}{5}$. You were close.