Q. 1.11 - A First Course in Probability

Question

In how many ways can $3$novels, $2$mathematics books, and $1$chemistry book be arranged on a bookshelf if

(a) the books can be arranged in any order?

(b) the mathematics books must be together and the novels must be together?

(c) the novels must be together, but the other books can be arranged in any order?

Step-by-Step Solution

Answer

(a) The books can be arranged in $720$ ways.

(b) The  books can be arranged in $72$ ways.

(c)  The books can be arranged in $144$ ways.

Part (a) Step 1. Given information.

There are $3$novels, $2$ mathematics books and $1$ chemistry book. These books are to be arranged on a bookshelf in any order.

Part (a) Step 2. Find the number of ways.

Number of arrangements = $\left(3+2+1\right)! = 6! \text{ways}$

$$\begin{gathered} =6\times 5\times 4\times 3\times 2\times 1 \\ =720 \text{ways} \end{gathered}$$

Therefore, the books can be arranged in $720$ ways.

Part (b) Step 1. Given information.

There are $3$ novels, $2$ mathematics books and $1$ chemistry book. These books are to be arranged on a bookshelf such that the mathematics books must  be together and the novels must  be together.

Part (b) Step 4. Find the number of ways.

Number of ways in which novels can be arranged = $3!$

Number of ways in which mathematics books can be arranged = $2!$

Number of ways in which Chemistry book can be arranged = $1!$

These three books can be arranged in = $3!$ ways

So, total number of ways $=3!\times 2!\times 1!\times 3!$

$$\begin{gathered} =\left(3\times 2\times 1\right)\times \left(2\times 1\right)\times \left(1\right)\times \left(3\times 2\times 1\right) \\ =72 \text{ways} \end{gathered}$$

Therefore, the books can be arranged in $72$ ways.

Part (c) Step 1. Given information.

There are $3$ novels, $2$ mathematics books and $1$ chemistry book. These books are to be arranged on a bookshelf such that the novels must be together.

Part (c) Step 2. Find the number of ways.

Total novels comprise as one group.

So, total number of groups = $4$

Number of ways in which four groups can be arranged = $4!$

Number of ways in which novels can be arranged = $3!$

So, total number of ways

$$\begin{gathered} =4!\times 3! \\ =(4\times 3\times 2\times 1)\times (3\times 2\times 1) \\ =144 \end{gathered}$$

Therefore, the books can be arranged in $144$ ways.