Misc 5 (Method 1)
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
A number is divisible by 10
if 0 is at the units place
Thus, We need to form 6 digit number
whose unit place is 0
So,
We need to fill up 5 places with the remaining digits 1, 3, 5, 7, & 9
120 is divisible by 10
as last digit is 0
Hence, n = Number of digits = 5
& r = Number of places to fill = 5
Number of 6 digit numbers = 5P5
= 5!/(5 − 5)!
= 5!/0!
= 5!/1
= 5 × 4 × 3 × 2 × 1
= 120
Misc 5 (Method 2)
How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
A number is divisible by 10
if 0 is at the units place
Thus, We need to form 6 digit number
whose unit place is 0
So,
We need to fill up 5 places with the remaining digits 1, 3, 5, 7, & 9
120 is divisible by 10
as last digit is 0
Thus,
Number of 6 digit numbers = 1 × 5 × 4 × 3 × 2 × 1
= 120

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.