Question

Find the number of 3-digit numbers formed using the digits 1 to 9, without repetition, such the numbers either have all digits less than 5 or all digits greater than 4.

Answer 1

Answer: 120

Explanation:

The total number of digits from 1 to 9 = 10

The number of digits from less than 5 (0,1,2,3,4)=5

Since repetition is not allowed so we use Permutations , then the number of 3-digit different codes will be formed :-

`^5P_3=(5!)/((5-3)!)=(5*4*3*2!)/(2!)=5*4*3=60`

The number of digits from greater than 4 (5,6,7,8,9)=5

Similarly, Number of 3-digit different codes will be formed :-

`^5P_3=60`

Hence, the required number of 3-digit different codes = 60+60=120