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

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?

• ${6\choose 2}$ are the number of pairs. Once you pick a first pair on a list there are ${4\choose 2}$ pairs to put in the second position only a list and once you pick that there is ${2\choose 2}$ (0r $1$) pair to put in the last place on the list. SO there are ${6\choose 2}{4\choose 2}{2\choose 2}$ ways to make a list of three pairs. But the order of the list does not matter. So you divide it by the number of ways to order three items on a list. Commented Apr 9, 2019 at 23:00

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.