Nine people sit down at random seats around a round table. Four of them are math majors, three others are physics majors, and the two remaining

Nine people sit down at random seats around a round table. Four of them are math majors, three others are physics majors, and the two remaining people are biology majors. What is the probability that all four math majors sit in consecutive seats?

There are a total of $(9-1)!=8!$ possible seating arrangements. We treat the four math majors as a single person. So we're really seating 6 people around a table which gives $5!$. So the probability is $5!/8!$. I think this is wrong. Any help is greatly appreciated.

• Why do you "think this is wrong"? – Henry Nov 16 '17 at 1:16
• The probability just seems to be really small and I'm surprised it is that small, but I guess that doesn't really matter. I'm just not certain. It makes sense to me though. – ddswsd Nov 16 '17 at 1:22
• The "math person" can sit down in any of $4!=24$ ways. So you need to multiply by that factor. – Barry Cipra Nov 16 '17 at 1:23

There are $9$ distinct spots in which the $4$ math majors can be seated consecutively around a table, if you treat the $4$ people as a clump. There $4!$ ways to arrange them within each clump. The other $5$ people can be seated in $5!$ different ways. Everyone can be seated in $9!$ different ways giving a probability of
$$\frac{9\cdot4!\cdot5!}{9!}\approx.0714$$
Note that this probability is lower if you seat the $9$ people in a line rather than around a circle, since there would only be $6$ ways to "clump" the $4$ math majors together.
To see why there are $9$ ways to seat a clump of $4$ people, consider a clock with numbers $1-9$ instead of $1-12$. Then the $4$ people can sit in spots $$1-4, 2-5, 3-6, 4-7, 5-8, 6-9,7-1, 8-2, \text{ or } 9-3$$.
• Why do we multiply by $9$? – ddswsd Nov 16 '17 at 4:02
• @ddswsd: If you take the seats to be unnumbered, $\frac{4!5!}{8!}$ which amounts to the same thing. – true blue anil Nov 16 '17 at 4:35