Chapter 2, Problem 1E

To determine

To count: The number of four-digit numbers whose digits are 1, 2, 3, 4, or 5 and no property holds.

Answer to Problem 1E

The number of four digit numbers is 625.

Explanation of Solution

Procedure used:

Multiplication principle:

When a task has p outcomes and, no matter what the outcome of the first task, a second task has some q outcomes, then the two tasks performed consecutively will have $p\times q$ outcomes.

Calculation:

It is given that the four-digit number has to be formed from amongst the five digits.

As no property holds, each digit place of the four-digit number will be any of the five digits.

Hence, the number of four digit numbers becomes

$\begin{array}{c}\underset{\_}{5}\underset{\_}{5}\underset{\_}{5}\underset{\_}{5}=5\cdot 5\cdot 5\cdot 5\left(\begin{array}{l}\text{Every position in the number}\\ \text{has 5 choices}\end{array}\right)\\ =625\end{array}$

Therefore, the number of four digit numbers is 625.

To determine

To count: The number of four-digit numbers whose digits are 1, 2, 3, 4, or 5 and the property of the digits being distinct, holds.

Answer to Problem 1E

The number of four digit numbers is 120.

Explanation of Solution

Procedure used:

Multiplication principle:

When a task has p outcomes and, no matter what the outcome of the first task, a second task has some q outcomes, then the two tasks performed consecutively will have $p\times q$ outcomes.

Calculation:

It is given that the four-digit number has to be formed from amongst the five digits.

As property of the digit being distinct holds, each digit place of the four-digit number will be any of the five digits but one less from the previous in the next place.

Hence, the number of four digit numbers becomes

$\begin{array}{c}\underset{\_}{5}\underset{\_}{4}\underset{\_}{3}\underset{\_}{2}=5\cdot 4\cdot 3\cdot 2\left(\begin{array}{l}\text{Every position in the number}\\ \text{has one less choice from the previous}\end{array}\right)\\ =120\end{array}$

Therefore, the number of four digit numbers is 120.

To determine

To count: The number of four-digit numbers whose digits are 1, 2, 3, 4, or 5 and the property of the number being even.

Answer to Problem 1E

The number of four digit numbers is 250.

Explanation of Solution

Procedure used:

Multiplication principle:

When a task has p outcomes and, no matter what the outcome of the first task, a second task has some q outcomes, then the two tasks performed consecutively will have $p\times q$ outcomes.

Calculation:

It is given that the four-digit number has to be formed from amongst the five digits.

As the property of the number being even holds, each digit place of the four-digit number will be any of the five digits except at unit’s place.

The unit’s place of the numbers will carry even digits that are 2 and 4.

Hence, the number of four digit numbers becomes

$\begin{array}{c}\underset{\_}{5}\underset{\_}{5}\underset{\_}{5}\underset{\_}{2}=5\cdot 5\cdot 5\cdot 2\left(\begin{array}{l}\text{Every position in the number}\\ \text{has 5 choices except at unit's place}\end{array}\right)\\ =250\end{array}$

Therefore, the number of four digit numbers is 250.

To determine

To count: The number of four-digit numbers whose digits are 1, 2, 3, 4, or 5 and both the properties hold.

Answer to Problem 1E

The number of four digit numbers is 48.

Explanation of Solution

Procedure used:

Multiplication principle:

When a task has p outcomes and, no matter what the outcome of the first task, a second task has some q outcomes, then the two tasks performed consecutively will have $p\times q$ outcomes.

Calculation:

It is given that the four-digit number has to be formed from amongst the five digits.

As both the properties hold, each digit place of the four-digit number will be any of the five digits.

However, the choices for the unit’s place are 2 and the choices for other places starts from 4 and ends at 2.

Hence, the number of four digit numbers becomes

$\begin{array}{c}\underset{\_}{4}\underset{\_}{3}\underset{\_}{2}\underset{\_}{2}=4\cdot 3\cdot 2\cdot 2\left(\begin{array}{l}\text{Unit's place has 2 choices, other}\\ \text{places have }\le \left(5-1=4\right)\text{ choices}\\ \text{but different}\end{array}\right)\\ =48\end{array}$

Therefore, the number of four digit numbers is 48.