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?
Best Answer
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.