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Misc 5 - How many 6-digit numbers from 0, 1, 3, 5, 7, 9 - Miscellaneou

Misc 5 - Chapter 7 Class 11 Permutations and Combinations - Part 2

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Misc 5 - Chapter 7 Class 11 Permutations and Combinations - Part 3

Misc 5 - Chapter 7 Class 11 Permutations and Combinations - Part 4

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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

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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.