# No of ways in which a dozen people can be divided into one team of 2 and two teams of 5.

How many ways are there to split a dozen people into 3 teams, where one team has 2 people, and the other two teams have 5 people each?

I tried to answer the question and got the answer as $$\binom {12}{2} \binom{10}{5}$$.

But the solution I found out to be is $$\frac{\binom {12}{2} \binom{10}{5}}{2!}$$. 2! it seems was because I was forming two groups of 5 people.

So I thought of a simpler problem.

If I have four elements {A,B,C,D} and I want to split them in groups of two So there would $$\binom{4}{2} =6$$ ways of doing it.{A,B},{A,C}{A,D}{B,C}{B,D}{C,D}.But by the above logic I would have to divide the $$\binom {4}{2}$$ by 2!. But this doesn't seem correct.

Where is my understanding going wrong?

• It is correct to divide by $2!=2$ in the simpler problem, because if you think about it, $\{A,B\}$ and $\{C,D\}$(for example) give the same result of $\{\{A,B\},\{C,D\}\}$. We can form these pairs with all the possible sets, so it is correct to divide by $2$ in that situation. But if order does matter for the problem, then you don't divide by $2$, because then $\{\{A,B\},\{C,D\}\}$ and $\{\{C,D\},\{A,B\}\}$ would be different combinations, so we can't take out half the sets. Oct 19, 2020 at 7:37

$$(A,B)$$ and $$(C,D)$$
$$(A,C)$$ and $$(B,D)$$
$$(A,D)$$ and $$(B,C)$$
• So $\binom {4}{2}$ is just the no of ways of choosing two people from a group of 4. where as to find the no of ways in which two groups of two can found I divide by 2!. In the case of the earlier problem I was forming two groups of 5 people hence divide by 2!. So If I have n objects and wants to form k groups with r elements in each group I use $\frac {\binom{n}{r}}{k!}$ Oct 19, 2020 at 7:52
• Not exactly. Suppose you want to form $k$ groups all of size $r$ then the number of ways you can do this is counted as $$\frac{\binom{n}{r}\binom{n-r}{r}\ldots \binom{n-kr}{r}}{k!}$$ Oct 20, 2020 at 15:39