# I do not understand why do we divide with the k! that is used in nCk when multiplying two combinations or more combinations that has the same k value

I recently started studying Permutations and Combinations and following is one interesting question I came across.

There are 8 students in a class. The class teacher wants to divide divide those students into four teams. The sizes of teams need not all be equal and a team may consist of even one person. Show that the required team can be formed in 1701 ways.

How I approached the problem

We can select students in following 5 ways,

1. 5 Students, 1 Student, 1 Student, 1 Student
2. 4 Students, 2 Students, 1 Student, 1 Student
3. 3 Students, 3 Students, 1 Student, 1 Student
4. 3 Students, 2 Students, 2 Students, 1 Student
5. 2 Students, 2 Students, 2 Students, 2 Students

Then we can find combinations as below for the given five cases,

1. $$\binom{8}{5} \cdot \binom{3}{1} \cdot \binom{2}{1} \cdot \binom{1}{1}$$
2. $$\binom{8}{4} \cdot \binom{4}{2} \cdot \binom{2}{1} \cdot \binom{2}{1}$$
3. $$\binom{8}{3} \cdot \binom{5}{3} \cdot \binom{2}{1} \cdot \binom{2}{1}$$
4. $$\binom{8}{3} \cdot \binom{5}{2} \cdot \binom{3}{2} \cdot \binom{1}{1}$$
5. $$\binom{8}{2} \cdot \binom{8}{2} \cdot \binom{8}{2} \cdot \binom{8}{2}$$

## But above answer has been wrong and the correct answer is

1. $$\binom{8}{5}$$
2. $$\binom{8}{4} \cdot \binom{4}{2}$$
3. $$\binom{8}{3} \cdot \binom{5}{3}$$ divided by 2!
4. $$\binom{8}{3} \cdot \binom{5}{2} \cdot \binom{3}{2}$$ divided by 2!
5. $$\binom{8}{2} \cdot \binom{8}{2} \cdot \binom{8}{2} \cdot \binom{8}{2}$$ divided by 4!

Now I have 2 questions,

1. How come second one is correct? Why aren't we choosing in the scenarios where 1 student is present?
2. Why are we dividing with 2! and 4! etc... in second secnario?

More importantly how to recognise when to divide with 2! and 4! etc in problems...?

• Well, it seems as if the teams aren't named in any way. Hence it makes no difference if you permute the $1-$member teams. Indeed, it makes no difference if you permute any teams that have the same number of members.
– lulu
Commented Sep 26, 2023 at 19:17
• Your answer would have been correct if the problem had said: There are four teams, $A,B,C,D$ which need to be populated and $8$ available people. But that's a different question.
– lulu
Commented Sep 26, 2023 at 19:19
• If you are selecting students for several teams of the same size, notice that the different teams can be interchanged, and this shouldn't be considered a different solution. So just like you divide by $k!$ to account for different ways of ordering individual students, you also need to account for different ordering of teams. Commented Sep 26, 2023 at 19:19

Take case $$3$$ with two teams of $$3$$ Suppose people $$A,C,F$$ are in one team and $$B,D,X$$ in the other team. Now you could interchange them to $$B,D,X = ACF$$ and it is the same two teams, so we need to divide by $$2!$$
Similarly, if $$3$$ teams have the same size, we need to divide by $$3!$$