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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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