Six people, call them a, b, c, d, e, and f, are randomly divided into three groups of two. find the probability that a and b are in the same group, as are c and d. (do not impose unwanted ordering among groups.)

First lets find the number of permutations of three ordered groups, A and B, C and D and E and F:
3P3 = 6
However when the order in each of the pairs can be reversed, the number of arrangements meeting the required grouping becomes:
6 * 8 = 48.
The total number of permutations of the 6 people is given by:
6P6 = 720.
Therefore the required probability is found from:

+1,000)=g7e=mc2-g=7 because if you know algebra you should know g=7 to get your answer