91影视

It is given that, there are three classes and each class has $n$number of students and we have to select $3$students from the total number of students.

To select$3$students from$3n$students, the possible no. of choices arelocalid="1649084012796" $\left(\begin{array}{c}3n\\ 3\end{array}\right)$.

No. of ways of selecting $3$students from a class of $n$students is role="math" localid="1649055941717" $=\left(\begin{array}{c}n\\ 3\end{array}\right)$.

No. of ways of choosing any of the $3$classes $=\left(\begin{array}{c}3\\ 1\end{array}\right)=3$.

Therefore, the possible no. of choices that all $3$ students are in the same class are$3\left(\begin{array}{c}n\\ 3\end{array}\right)$.

No. of ways of selecting $2$students from a class of $n$students is $=\left(\begin{array}{c}n\\ 2\end{array}\right)$.

No. of ways of selecting $1$student from a class of $n$students is $=\left(\begin{array}{c}n\\ 1\end{array}\right)$.

The class from which $2$students are selected can be chosen in $\left(\begin{array}{c}3\\ 1\end{array}\right)\text{ways}=3$

The class from which $1$students is selected can be chosen in $\left(\begin{array}{c}2\\ 1\end{array}\right)\text{ways}=2$

Therefore, the possible no. of ways are$=3脳\left(\begin{array}{c}n\\ 2\end{array}\right)脳2脳\left(\begin{array}{c}n\\ 1\end{array}\right)$

No. of ways of selecting$1$student from $n$students is $\left(\begin{array}{c}n\\ 1\end{array}\right)=n$

No. of choices for $3$students $=n脳n脳n$

Therefore, the possible no. of choices that all 3 students are in different classes$=n^3$.

Using the results of parts (a) through (d), the combinatorial identity will be

$\left(\begin{array}{c}3n\\ 3\end{array}\right)=3\left(\begin{array}{c}n\\ 3\end{array}\right)+3\left(\begin{array}{c}n\\ 2\end{array}\right)脳2\left(\begin{array}{c}n\\ 1\end{array}\right)+n^3$