In how many ways can 6 children be divided into 3 pairs if the order of pairs does not matter?

Question

In how many ways can $6$ children be divided into $3$ pairs if the order of pairs does not matter?

In my book, the answer is only :

$$\frac{^6C_2\cdot^4C_2\cdot^2C_2}{3!}=\frac{1}{6}\cdot90=15$$

I do not understand why we divide it by $3!$ . What's the intuition behind this?

Answer 1

Cause you are choosing an order in the groups. If i have the children as $x_1x_2\dots x_6$ then if we pick $2$(by the $\binom{6}{2}$) we can color it with $\color{red}{red}$ for example $$\color{red}{x_1}x_2\color{red}{x_3}x_4x_5x_6,$$ then we pick other two and color them with blue, for example $$\color{red}{x_1}\color{blue}{x_2}\color{red}{x_3}x_4\color{blue}{x_5}x_6.$$ This is equivalent to $$\color{blue}{x_1}\color{red}{x_2}\color{blue}{x_3}x_4\color{red}{x_5}x_6$$ and any permutation of the $3$ colors. To avoid this you divide by the $3!$order of the colors.

Answer 2

Was going through a similar question and found a visual aid that helped me picture the question. Imagine a 6x6 table, where it is labeled 1 thru 6 horizontally and vertically as such.

Next we pair them up:

Since we cannot pair two of the same person, hence, we will take out 11, 22, ... 66; which are the diagonal pairs highlighted in yellow

Lastly, we take out half the triangle which double counted AB, BA, etc... highlighted in green. This also interprets 2 ways (e.g.: AB or BA) of choosing. And yes this is pascals triangle. This visual aid is only applicable when there is equal distribution.

Mathematically; we start with 6 x 6 = 36, next we subtract the yellow highlighted pairs 36 - 6 = 30. Lastly we subtract the green highlighted pairs 30 - 15 = 15.