Chapter 15.5, Problem 12WE

To determine

To calculate: The number of positive odd integers less than 10,000 with digits 3, 4, 6, 8 and 0.

Answer to Problem 12WE

The number of positive odd integers less than 10,000 with digits 3, 4, 6, 8 and 0are $125$ .

Explanation of Solution

Given information:

Positive odd integers less than 10,000 are to be formed with digits 3, 4, 6, 8 and 0.

Formula used:

Fundamental counting principle states that if there are m ways to make a selection and n ways to make a second selection then there are $m\times n$ number of ways in which two selection can be made.

Calculation:

Consider the provided information that positive odd integers less than 10,000 are to be formed with digits 3, 4, 6, 8 and 0.

Out of the provided digits if the number ends with 3 then only it is an odd number.

Case I: The number of one digit numbers.

Let 1-digt number be $\boxed{}$ .

Out of the digits provided only one way to fill the box that is with digit 3 $\boxed{1}$ .

Therefore, there is 1 one digit odd integer.

Case II: The number of two digit numbers.

Let 2-digit number be $\boxed{}\boxed{}$ .

Out of the digits provided only one way to fill the last box that is with digit 3, $\boxed{}\boxed{1}$ and any one of the 4 digits can be filled in it as 0 cannot be filled in either of the box $\boxed{4}\boxed{1}$ .

Therefore, number of ways are.

  $\text{4}\times \text{1=4}$

Case III: The number of three digit numbers.

Let 3-digit number be $\boxed{}\boxed{}\boxed{}$ .

Out of the digits provided only one way to fill the last box that is with digit 3, $\boxed{}\boxed{}\boxed{1}$ and any one of the 5 digits can be filled in second box $\boxed{}\boxed{5}\boxed{1}$ and any one of the 4 digits can be filled in first box as 0 cannot be filled in the box $\boxed{4}\boxed{5}\boxed{1}$ .

Therefore, number of ways are.

  $\text{4}\times 5\times \text{1=20}$

Case III: The number of four digit numbers.

Let 4-digit number be $\boxed{}\boxed{}\boxed{}\boxed{}$ .

Out of the digits provided only one way to fill the last box that is with digit 3, $\boxed{}\boxed{}\boxed{}\boxed{1}$ and any one of the 5 digits can be filled in second box and third box $\boxed{}\boxed{5}\boxed{5}\boxed{1}$ and any one of the 4 digits can be filled in first box as 0 cannot be filled in the box $\boxed{4}\boxed{5}\boxed{5}\boxed{1}$ .

Therefore, number of ways are.

  $\text{4}\times 5\times 5\times \text{1=100}$

Recall that Fundamental counting principle states that if there are m ways to make a selection and n ways to make a second selection then there are $m\times n$ number of ways in which two selection can be made.

Therefore, the number of ways to select are,

  $100+20+4+1=125$

Thus, number of positive odd integers less than 10,000 with digits 3, 4, 6, 8 and 0 are $125$ .