Find probability of having a female in every group 10 male students and 5 female students are split into 5 groups (every grup consists of 3 students). What is the probability that there is a female student in every group?
I should solve this with combinatorics. Any idea on how to appraoch the problem?
 A: Hint: Suppose you have 15 chairs of 5 different colors.
How many ways are there to choose 5 chairs for the women to sit on?
How many of those ways use exactly one chair of each color?
A: There are $\frac{15!}{3!^5}$ possibilities to distribute the students.
If every group has a female student, there are 
$\frac{10!}{2!^5}$ possibilities to distribute the male students.
Dividing the two numbers gives $\frac{27}{40040}=0.000674$
A: You may find it much simpler to consider just placing a woman in each group.
The first woman has to go to one group or another.  
The second woman now has $12$ permissible slots out of $14,$ the third, $9$ slots out of $13$, and so on
Thus $ Pr = \dfrac{12}{14}\cdot\dfrac{
9}{13}\cdot\dfrac6{12}\cdot\dfrac3{11}= \dfrac{81}{1001}$
PS
Since my answer is differing from the accepted one, here is confirmation by a more conventional approach, conceptually simple, even if involving more computations:
$Pr = \dfrac{\dbinom{5}{1,1,1,1,1}\dbinom{10}{2,2,2,2,2}}{\dbinom{15}{3,3,3,3,3}}$
PPS
@craaaft: See here
A: The total number of ways of splitting the groups is (14C2)(11C2)(8C2)(5C2)(2C2)= 13*7*11*5*7*4*5*2=280*55*91=1401400. The given case has these many ways= (10C2)(8C2)(6C2)(4C2)(2C2)= 45*28*15*6=45*90*28=113400. Therefore, the probability is 0.0809.
I have observed that Mr. Peter has put up a solution that seems different, but also wish to add that I believe that his solution overlooks the fact that the groups can be numbered in any order, that is the order of the groups does not matter, so his total number of ways of splitting groups should be 15!/{[(3!)^5]*5!}. If this is done, then his probability will be 120*0.000674=0.08088.
A: total possible: ${15\choose 3}{12\choose 3}{9\choose 3}{6\choose 3}{3\choose 3}$
one female in every group:  ${10\choose 2}{5\choose 1}{8\choose 2}{4\choose 1}{6\choose 2}{3\choose 1}{4\choose 2}{2\choose 1}{2\choose 2}{1\choose 1}$
solution: ${{10\choose 2}{5\choose 1}{8\choose 2}{4\choose 1}{6\choose 2}{3\choose 1}{4\choose 2}{2\choose 1}{2\choose 2}{1\choose 1}}\over{{15\choose 3}{12\choose 3}{9\choose 3}{6\choose 3}{3\choose 3}}$
