# 60 students sitting at 5 tables randomly

There are $$60$$ students from 12 different grade years, $$5$$ students from each year.

Also, there are $$5$$ tables A,B,C,D,E. Tables A-D have $$14$$ seats and E has $$4$$ . Students sit randomly to the Tables.

What is the probability that all students from a specific grade year(5-students) sit at table A.

I was thinking it is a classical probability with maybe hyper-geometric form. something like: $$\frac{\binom{5}{5}*\binom{55}{9}}{\binom{60}{14}}$$ but I am not sure if i have to consider the other tables as well, since students sit randomly?

• You don't have to consider other tables for this question. Think: is there any difference if there are no extra tables, one extra table, or two extra tables? No, all the other students will not be sitting at the first table. – Quang Hoang May 21 at 11:54
• Since you have determined table A, so your respond is correct. If you want to change sitting problem, you must add more details, like : probability of sitting all student from a grade on one table, or probability of sitting all student from at least one grade on one table ,... – BarzanHayati May 21 at 11:57

Let us label the grade years with $$i=1,\dots12$$ and let $$E_i$$ denote the event that the students of grade $$i$$ are all seated at table A.

Then to be found is: $$P\left(\bigcup_{i=1}^{12}E_i\right)$$

For this we can use the principle of inclusion/exclusion together with symmetry based on the fact that the events $$E_i$$ are evidently equiprobable.

This results in:$$P\left(\bigcup_{i=1}^{12}E_i\right)=\binom{12}1P(E_1)-\binom{12}2P(E_1\cap E_2)=\frac{\binom{12}1\binom{14}5}{\binom{60}5}-\frac{\binom{12}2\binom{14}{10}}{\binom{60}{10}}$$

Just counting friend:

Total number of different seat combination: X = 60!

Total number of combination, which only one specific grade sits at A: $$Y= \binom{14}5 * 55! - 11 * \binom{9}5*50!$$

Total number of combination, which only two specific grades sit at A: $$Z=\binom{12}2 * \binom{14}{10}*50!$$

Probability = (Y+Z)/X