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

Step-by-step 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 :-

$^5P_3=\dfrac{5!}{(5-3)!}=\dfrac{5\times4\times3\times2!}{2!}=5\times4\times3=60$

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