Number of possible permutations of three pairs of socks, one blue, one black, and one white? There are 3 colours of socks; blue, black and white and each colour has two pairs of socks and we need to distribute them to six people and the left leg socks are distinguishable from right leg socks. What are the possible ways of distribution?
Approach: align the right leg socks in a line in a random manner, the number of ways of distributing the left leg socks to this one particular distribution will be:
$$\frac{6!}{(2!)^3}$$
But this is the one way of distribution, What are the other possible ways?
Does squaring of the expression makes any sense, where order matters?


There are three pairs of socks, one black, one blue, and one white.  In how many ways can the socks be permuted if left socks are distinguishable from right socks?

Since there are six different socks, they can be arranged in $6!$ orders.

There are three pairs of socks, one black, one blue, and one white.  In how many ways can the socks be permuted if left socks are indistinguishable from right socks?

Choose two of the six positions for the black socks, two of the remaining positions for the blue socks, and fill the final two positions with white socks, which can be done in
$$\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{6!}{4!2!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{2!0!} = \frac{6!}{2!2!2!}$$
distinguishable ways.  This is what you counted.

There are three pairs of socks, one black, one blue, and one white.  In how many ways can pairs of socks be permuted if left socks are indistinguishable from right socks?

There are three pairs of socks, which are distinguishable by their colors.  Hence, there are $3!$ possible arrangements.

There are three pairs of socks, one black, one blue, and one white.  In how many ways can pairs of socks be permuted if left socks are distinguishable from right socks?

There are three pairs of socks, which are distinguishable by their colors.  There are $3!$ ways to arrange the pairs.  Each pair can be arranged in $2!$ ways.  Hence, there are $3!(2!)^3$ possible arrangements.
