Combination small challenge problem How many ways can 3 different Scientific Groups be formed using 5 students such that
Each student is at least be a member of one committee and
each two committee has exactly 2 students in common?
I tried this solution:
$$\left(\binom{5}{4}×4!\times\binom{3}{1}=360\right) + \left(\binom{5}{2}+\binom{5}{3}\right) = 460$$
But I think I'm'm wrong. Any ideas? Thanks.
 A: EDIT: I had originally solved the question assuming that each committee had exactly three members. The solution in that case is below.
Call a student diligent if he or she sits on all three committees. Then there must be exactly one or two diligent students. If there are two, $A$ and $B$, then either each committee has three members, and so they are of the form $\{ABC,ABD,ABE\}$ (see original post below); or one committee has four members, one has three, and one has two, and they are of the form $\{ABCD,ABE,AB\}$. In this case there are $\binom{3}{2}=3$ ways to choose who will form the committee of size four, at which point the other two committees are determined. It cannot be the case that one committee has all five members, for then the remaining two committees would be restricted to two members, but would then be identical.
Thus, so far we have $\binom{5}{2} (1 + 3) = 40$ ways to form committees with two diligent students.
If there is one diligent student $A$, then there are three students ($B,C,D$) who each sit on two committees, and one, $E$, who only sits on one; the committees are of the form $\{ABCE,ABD,ACD\}$. There are five ways to choose the diligent student,  $\binom{4}{3} = 4$ ways to choose the almost-as-diligent, and a choice of three committees for the final student to join.
This gives $5\cdot4\cdot3=60$ arrangements with one diligent student.
Total: $40 + 60 = 100$.
Finally, if the committees are labelled—hopefully they each have some name and purpose—then there are $100 \cdot 3! = \mathbf{600}$ ways.

Original solution (when each committee has three members):

Suppose that students $A,B,C$ form the first committee, and $A,B,D$ form the second (two of the students must be common between these two committees). Then, since every student must appear on at least one committee, the final committee must include the fifth student, $E$. The other two students on this third committee must be $A$ and $B$, since any other pair would not meet the requirement that any pair of committees have two members in common.
So, you can see that there are two diligent students, $A$ and $B$, who must sit on each of the three committees, while the others, $C$, $D$, and $E$, each sit on only one.
There are $\binom{5}{2}$ ways to choose two diligent students from five.

Edit: Let me also add the following observation as an example of how to check to see whether your initial calculation was reasonable. Let's forget about the restrictions for the moment and simply calculate how many ways there are to form three distinct committees of three members each. There are $\binom{5}{3} = 10$ different such committees, so there are $$\binom{\binom{5}{3}}{3} = \binom{10}{3} = 120$$ different ways to form three committees. Your answer must therefore be (much) less than $120$, so indeed, $460$ is off the mark.

A: We assume that the groups are distinguishable (since the problem states that they are different), so designate the groups by 1,2,3 and the students by A,B,C,D,E.
1) There are $\binom{5}{2}$ ways to choose the 2 students common to groups 1 and 2. (Assume they are A and B.)
2) Now consider the students common to groups 2 and 3.
$\;\;\;\;$a) If these are also students A and B, then each of the other 3 students
is in exactly one of the 3 $\;\;\;\;\;\;\;$groups; so there are $\binom{5}{2}\cdot3^3=270$ possibilities in this case. 
$\;\;\;\;\;\;\;$(Notice that some groups are identical in this case.)
$\;\;\;\;$b) If the students common to groups 2 and 3 are two of C, D, and E, then groups 1 and 3 cannot have $\;\;\;\;\;\;$ two members in common (since these two students are not in group 1 and A and B are not in group $\;\;\;\;\;\;\;$3), so this case is impossible. 
$\;\;\;\;$c) If the students common to groups 2 and 3 include one of A or B and one of C, D, and E, then there $\;\;\;\;\;\;\;$are $2\cdot3$ ways to choose these students. (Assume that they are A and C.)  
$\;\;\;\;\;\;$Since A is a common member of groups 1 and 3, the other common member is either D or E.
$\;\;\;\;\;\;$If D is the other common member, say, then E has to be in exactly one of the 3 groups, so there are
$\;\;\;\;\;\;\binom{5}{2}\cdot2\cdot3\cdot2\cdot3=360$ possibilities in this case.
Therefore there are a total of $270+360=630$ possible groups.
