Question

Question 4 How many integer numbers between 0 and 9999 are there that have exactly one digit 1 and exactly one digit 3

Answer 1

Answer:

768 numbers

Step-by-step explanation:

Given

$Digits = \{0,1,2,3,4,5,6,7,8,9\}$

Required

Number of digits between 0 ad 9999 that have one 1 and one 3

There are a total of 4 digits that make up any of the numbers in 0 to 9999 with the given condition

This can be represented as WXYZ

• Digit 1 can occupy any of the 4 positions

• Digit 3 can occupy any of the 4 - 1 positions

The remaining 8 digits will occupy the last 2 positions in the following ways:

• The first of the 8 digits can be selected from any of $\{0,2,4,5,6,7,8,9\}$ i.e. 8 ways

• The second can be selected from any of $\{0,2,4,5,6,7,8,9\}$ i.e. 8 ways

So, the total number of selection is:

$Total = 4 * (4 - 1) * 8 * 8$

$Total = 4 * 3 * 8 * 8$

$Total = 768$