Maximum Seating Chart Confusion: Students with Same Name ending up in Same Groups Help! I assigned my Statistics & Probability high school students a bonus problem that I can't solve! 
The problem: I have a class of 16 students, that I randomly assign to 4 table groups of size 4 each. In that class, there are two Ben's, two Katie's and two Sam's. Order within a table group doesn't matter, but table number does. 
A) What is the probability that a random seating chart will end in maximum confusion (each Ben/Katie/Sam assigned to the same table as the student with their same name)? 
B) What is the probability that there will be minimum confusion (no student assigned to a group with the student of their same name)?
I have figured out:
number(possible seating charts) $$= \binom {16} 4\cdot \binom {12} 4\cdot \binom 8 4\cdot \binom 4 4 = 63,063,000$$
n(seating charts with the two Ben's together)$$=4\cdot \binom 2 2\cdot \binom {14} 2\cdot \binom {12} 4\cdot \binom 8 4\cdot \binom 4 4= 12,612,600$$
(there are 4 tables they could sit at, "choose" the two of them, choose the remaining 2 people at the table, and then fill in the rest of the tables)
I'm pretty sure this works out correctly, with 3/15 = 1/5 chance that the second Ben is assigned to one of the 3 remaining open seats at the same table as the first Ben.
Where I'm stuck:
I could do the same process for Katie's and Sam's, but those 12 million double-Ben seating charts also include some double Katie and double Sam seating charts. So I would need to subtract those out (so I don't double-count them). But how many are there? More specifically, how many numbers will I need to subtract? The charts with BenBenKatieKatie at one table, and the charts with BenBen at one table and KatieKatie at a different table (which would be multiplied by 3 rather than 4, yes?), and the charts with BenBenSamSam (but some of those are also the BenBen + KatieKatie charts)...
I feel like there's got to be an easier way, but I don't see it.
And I'm at much more of a loss for part B. I can calculate the probability that Bens will be at different tables, but the double counting goes similarly bonkers in my head.
Thank you!
Laura
 A: You have correctly calculated the number of possible seating arrangements.

What is the probability that a random seating chart will result in maximum confusion (each Ben/Sam/Katie assigned to the same table as the other student with the same name)?

There are two possibilities:


*

*There is one pair each at three different tables.

*There are two pairs at one table and one pair at a second table.


Three different tables:
There are four ways to assign a table to the two Ben's, three ways to assign one of the remaining tables to the two Katie's, and two ways to assign one of the remaining tables to the two Sam's.  This leaves ten students whose seats have not been assigned.  There are $\binom{10}{2}$ ways to assign two of them to sit with the Ben's, $\binom{8}{2}$ ways to assign two of the remaining eight students to sit with the two Katie's, and $\binom{6}{2}$ ways to assign two of the remaining six students to sit with the two Sam's.  The remaining four students must sit at the remaining table.
$$4 \cdot 3 \cdot 2 \cdot \binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{4}$$
Two different tables:
There are four ways to choose the table which will receive two pairs.  There are $\binom{3}{2}$ ways to choose two pairs to sit at that table.  There are three ways to choose the table that will receive the remaining pair.  That leaves ten students whose seats who have not been assigned.  There are $\binom{10}{2}$ ways to choose the two students who will sit with the single pair.  There are $\binom{8}{4}$ ways to choose which four students will sit at the lowest numbered remaining table.  The remaining four students must be seated at the remaining table.
$$4 \cdot \binom{3}{2} \cdot 3 \cdot \binom{10}{2} \binom{8}{4} \binom{4}{4}$$  
Since these two cases are mutually disjoint, the probability of maximum confusion is 
$$\frac{4 \cdot 3 \cdot 2 \dbinom{10}{2}\dbinom{8}{2}\dbinom{6}{2}\dbinom{4}{4} + 4 \cdot \dbinom{3}{2} \cdot 3 \dbinom{10}{2}\dbinom{8}{4}\dbinom{4}{4}}{\dbinom{16}{4}\dbinom{12}{4}\dbinom{8}{4}\dbinom{4}{4}}$$

What is the probability that a random seating chart will result in minimum confusion (no student assigned to a table with a student of the same name)?

For the numerator, subtract the number of seating arrangements in which at least one pair of students sit at the table.  To do so, you will need to apply the Inclusion-Exclusion Principle.
