combinatorics difficult problem? 
"A group of 10 married couples (each consisting of one man and one woman) is divided in 5 groups of 4 people in each one. How many ways are there to divide the 10 couples so that each boat (there are 5 boats) contains 2 men and 2 women?"

First, I counted all the possible ways to arrange the men in the boats. I used combinations and I found that in the first boat there are $(10 \times 9)/2$ combinations of men, in the second boat there are $(8 \times 7) /2$ combinations of men, in the third boat there are $(6 \times 5) /2$ combinations of men, in the forth boat: $(4 \times 3)/2$ combinations and in the fifth boat $(2 \times 1)/2$ combinations of men, but my teacher said that I am counting more combinations of which there are, and he also told me that I need to divide by $5$! but I don't know why I need to do this; I would appreciate your help.
 A: Let's think of just four men and two boats. By your argument there should be ${4 \choose 2}=6$ ways to assign them. If the men are A,B,C,D the splits are AB-CD,AC-BD,AD-BC,BC-AD,BD-AC,CD-AB. But AB-CD is the same as CD-AB because the boats are interchangeable, so you need to divide by $2!$. Similarly in the original problem you can order the boats in $5!$ ways, so need to divide by that.
A: Consider two boats, with 4 men (M1, M2, M3, M4):
Boat 1: M1 M2 and Boat 2: M3 M4 is equivalent to Boat 1: M3 M4 and Boat 2: M1 M2.
This is the same as the equivalence of Boat 1: M2 M1 and Boat 2: M2 M1
A: Number of ways how men can arranged = 10C2.8C2.6C2.4C2.2C2/(5!)... Since, for each set of groups of men chosen given by only the numerators undergoes permutations... There are 5 groups with 2 men in each, hence for only one set the number of permutation is 5!... Now, number of ways how women are chosen for each set of men group chosen = 10C2.8C2.6C2.4C2.2C2... Thus, total required number of ways = (10C2.8C2.6C2.4C2.2C2)^2/(5!)... Is my answer useful..???...
