Q. 1.9 - A First Course in Probability

Question

Consider three classes, each consisting of $n$ students. From this group of $3n$ students, a group of $3$ students is to be chosen.

(a) How many choices are possible?

(b) How many choices are there in which all $3$ students are in the same class?

(c) How many choices are there in which $2$ of the $3$ students are in the same class and the other student is in a different class?

(d) How many choices are there in which all $3$ students are in different classes?

(e) Using the results of parts (a) through (d), write a combinatorial identity.

Step-by-Step Solution

Answer

(a) The possible no. of choices are $\left(\begin{array}{c}3n\\ 3\end{array}\right)$.

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

(c) The possible no. of choices that $2$ of the $3$ students are in the same class and the other student is in a different class are  $3\times \left(\begin{array}{c}n\\ 2\end{array}\right)\times 2\times \left(\begin{array}{c}n\\ 1\end{array}\right)$.

(d) The possible no. of choices that all $3$ students are in different classes are $n^3$

(e) Using the results of parts (a) through (d), the combinatorial identity will be  $\left(\begin{array}{c}3n\\ 3\end{array}\right)=3\left(\begin{array}{c}n\\ 3\end{array}\right)+3\left(\begin{array}{c}n\\ 2\end{array}\right)\times 2\left(\begin{array}{c}n\\ 1\end{array}\right)+n^3$

Part (a) Step 1. Given information.

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.

Part (a) Step 2. Find the possible no. of choices.

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

Part (b) Step 1. Find the possible no. of choices that all 3 students are in the same class.

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

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 $3\left(\begin{array}{c}n\\ 3\end{array}\right)$.

Part (c) Step 1. Find the possible no. of choices that 2 of the 3 students are in the same class and the other student is in a different class.

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

Part (d) Step 1. Find the possible no. of choices that all 3 students are in different classes.

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 $=n\times n\times n$

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

Part (e) Step 1. Write a combinatorial identity.

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

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