The teacher has 25 students in her class. She takes 5 of them at a time, to a park as often she can, without taking the same 5 kids more than once. Find the number of visits, the teacher makes to the park and also the number of visits every kid makes.
Attempt
So, I tried to do this and came to a thought that this question means that the same lot can't go to park twice or more. But, any number of students less than 5 can go more than 1 time being a different lot. But, I facing problem on accounting this. It is coming to be very big number which is I think is absurd.
Also the answer key has the answers $25_{C_5}$ & $24_{C_4}$ respectively which is very less as compared to my answer which is more than 7 digit figure.
Let's consider a smaller example. Suppose the teacher has six students, $A, B, C, D, E, F$, and takes three students to the park at the time. There are $\binom{6}{3} = 20$ groups she can take. They are $$ \begin{array}{c c} \color{red}{A}, B, C & B, C, D\\ \color{red}{A}, B, D & B, C, E\\ \color{red}{A}, B, E & B, C, F\\ \color{red}{A}, B, F & B, D, E\\ \color{red}{A}, C, D & B, D, F\\ \color{red}{A}, C, E & B, E, F\\ \color{red}{A}, C, F & C, D, E\\ \color{red}{A}, D, E & C, D, F\\ \color{red}{A}, D, F & C, E, F\\ \color{red}{A}, E, F & D, E, F \end{array} $$ Notice that $A$ is in $\binom{5}{2} = 10$ of these groups, once for each of the $\binom{5}{2}$ ways the teacher can select two students to go to the park with $A$.
In your problem, there are $\binom{25}{5}$ ways for the teacher to select $5$ of the $25$ students to go to the park and $\binom{24}{4}$ ways for the teacher to select four other students to go the park with a particular student. Hence, each student is in $\binom{24}{4}$ of the $\binom{25}{5}$ possible groups the teacher can take to the park.