12 distinct pairs of gloves combinatorics Given six distinct pairs of gloves,  12 distinct gloves in all,  how many ways are there to distribute two gloves to each of six sisters? 
a)  if the two gloves someone receives might both be for the left hand or right hand. 
b) if each sister gets one left and one right. 
Soln:
a)  So i thought about dissectng the problem first into finding the number of ways in which I could make pairs of gloves:   
C(12, 2)
then I asked how many ways could the sets of gloves be assigned amongst the 6 sisters:
$6!$
and taking the product of this would give me the amount of ways this could occur:
= C(12,2) * 6!
then for part b) 
using this same idea except:
C(6,1) C(6,1) for the left and right glove each .
the  C(6,1)*C(6,1)*6!
Right way to think them through? 
 A: Some components in the proposed answers are relevant, but the answers are at quite a distance from the truth.
a) Line up the people in order of age. The gloves for the leftmost one can be chosen in $\binom{12}{2}$ ways. For every such way, the gloves for the next person can be chosen in $\binom{10}{2}$ ways. For every way of assigning gloves to the first two people, there are $\binom{8}{2}$ ways to assign gloves to the third. And so on, for a total of $\binom{12}{2}\cdot\binom{10}{2}\cdot\binom{8}{2}\cdots\binom{2}{2}$.
This expression can be greatly simplified, to $\frac{12!}{2^6}$. You may want to look for an argument that yields this answer directly.
b) There are $(6)(6)$ ways to assign right and left gloves to the leftmost person. For every such way, there are $(5)(5)$ ways to assign right and left gloves to the next person, and so on.
Or else imagine distributing the right gloves first. This can be done in $6!$ ways. For each way, the left gloves can be distributed in $6!$ ways, for a total of $(6!)^2$.
