Last updated at Dec. 24, 2019 by Teachoo

Transcript

Example 11 (Method 1) How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed? Number between 100 and 1000 will be 3 digit number 3 digit number will be formed from the digits 0, 1, 2, 3, 4, 5 But, these include numbers starting with ‘0’ like 051, 012, …etc which are actually 2 digit numbers Required numbers = Total 3 digit numbers – 3 digit number which have 0 in the beginning Total 3 digit numbers Total digits in 0, 1, 2, 3, 4, 5 = 6 n = 6 Here, we have to form 3-digit number, So, r = 3 Total number of 3 digits number = nPr = 6P3 = 6!/(6 − 3)! = 6!/3! = (6 × 5 × 4 × 3!)/3! = 6 × 5 × 4 = 120 Finding numbers that start with 0 Hence number of digit left (exclude 0 from 0, 1, 2, 3, 4, 5) = 5 So n = 5 Number of digit required = 2 r = 2 Total numbers of 3 digit number starting with ‘0’ = nPr = 5P2 = 5!/(5 − 2)! = 5!/3! = (5 × 4 × 3!)/3! = 5 × 4 = 20 Hence The required number is = Total 3 digit numbers – 3 digit number which have 0 in the beginning = 120 – 20 = 100 Example 11 (Method 2) How many numbers lying between 100 and 1000 can be formed with the digits 0, 1, 2, 3, 4, 5, if the repetition of the digits is not allowed? Number between 100 and 1000 will be 3 digit number Let the 3 digit number be If 0 is in hundred’s place, the number will be like 023, 056,etc which is a two-digit number. So, 0 is not possible in hundreds place. Number of 3 digit numbers = 5 × 5 × 4 = 100

Examples

Example 1

Example 2

Example 3

Example 4 Important

Example 5

Example 6

Example 7

Example 8 Important

Example 9

Example 10

Example 11 Important You are here

Example 12 Important

Example 13

Example 14 Important

Example 15

Example 16 Important

Example 17

Example 18

Example 19 Important

Example 20 Important

Example 21 Important

Example 22 Important

Example 23 Important

Example 24 Important

Chapter 7 Class 11 Permutations and Combinations

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.