Question

How many integers are there that consist of four digits (from 1000 to 9999) where the first digit is a 5 or an 8 and the third digit is not 5 or 8

Answer 1

1600 integers

Explanation:

Since we have a four digit number, there are four digit placements.

For the first digit, since there can either be a 5 or an 8, we have the arrangement as ²P₁ = 2 ways.

For the second digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

For the third digit, since it neither be a 5 or an 8, we have two less digit from the total of ten digits which is 10 - 2 = 8. So, the number of ways of arranging that is ⁸P₁ = 8.

For the last digit, we have ten numbers to choose from, so we have ¹⁰P₁ = 10.

So, the number of integers that can be formed are 2 × 10 × 8 × 10 = 20 × 80 = 1600 integers