Misc 5 - Chapter 6 Class 11 Permutations and Combinations

Transcript

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