Last updated at March 8, 2017 by Teachoo

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Example 23 (Method 1) How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4? There are total 7 digits in 1000000 We need to form a 7 digit number using the digits 1, 2, 0, 2, 4,2, 4 But, these include numbers starting with β0β like 0412224, β¦etc which are actually 6 digit numbers Hence, we cant have number beginning with 0 So the number can begin either with 1, 2 or 4 Required numbers = Numbers starting with 1 + Numbers starting with 2 + Numbers starting with 4 Case 1 If number begin with 1 The remaining digit to be arranged will be 0, 2, 2, 2, 4, 4 Here three 2s & two 4s Since digit are repeating hence we use this formula = π!/π1!π2!π3! Number of remaining digit = 6 Thus, , n = 6 Since, three 2s & two 4s & p1 = 3, p2 = 2 The numbers of numbers beginning with 1 = π!/π1!π2! = 6!/(3! 2!) = (6 Γ 5 Γ 4 Γ 3!)/(3! Γ 2 Γ 1) = 60 Case 2 If number begin with 2 The remaining digit to be arranged will be 1,0, 2, 2, 4, 4 Here two 2s & two 4s Since digit are repeating hence we use this formula = π!/π1!π2!π3! Number of remaining digit = 6 Thus, , n = 6 Since, two 2s & two 4s & p1 = 2, p2 = 2 The numbers of numbers beginning with 2 = π!/π1!π2! = 6!/(2! 2! ) = (6 Γ 5 Γ 4 Γ 3 Γ 2 Γ 1)/((2 Γ 1) Γ (2 Γ 1) ) = 180 Case 3 If number begin with 4 The remaining digit to be arranged will be 1,0, 2, 2, 2, 4 Here three 2s Since digit are repeating hence we use this formula = π!/π1!π2!π3! Number of remaining digit = 6 Thus, , n = 6 Since, three 2s & p1 = 3 The numbers of numbers beginning with 4 = π!/(π1! ) = 6!/3! = (6 Γ 5 Γ 4 Γ 3!)/3! = 120 Required numbers = Numbers starting with 1 + Numbers starting with 2 + Numbers starting with 4 = 60 + 180 +120 = 360 Example 23 (Method 2) How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4? There are total 7 digits in 1000000 We need to form a 7 digit number using the digits 1, 2, 0, 2, 4,2, 4 But, these include numbers starting with β0β like 0412224, β¦etc which are actually 6 digit numbers Hence, we cant have number beginning with 0 Required numbers = All arrangements β Numbers starting with 0 All arrangements The digits to be arranged are 1, 0, 2, 2, 2, 4, 4 Here, three 2s & two 4s Since digit are repeating hence we use this formula = π!/π1!π2!π3! Number of digits = 7 Thus, , n = 7 Since, three 2s & two 4s & p1 = 3, p2 = 2 All arrangements = π!/π1!π2! = 7!/3!2! = (7 Γ 6 Γ 5 Γ 4 Γ 3!)/(3! Γ 2 Γ 1) = 420 If number begin with 0 The remaining digit to be arranged will be 1, 2, 2, 2, 4, 4 Here three 2s & two 4s Since digit are repeating hence we use this formula = π!/π1!π2!π3! Number of remaining digit = 6 Thus, , n = 6 Since, three 2s & two 4s & p1 = 3, p2 = 2 The numbers of numbers beginning with 1 = π!/π1!π2! = 6!/(3! 2!) = (6 Γ 5 Γ 4 Γ 3!)/(3! Γ 2 Γ 1) = 60

Ex 7.1,6
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Ex 7.1,4 Important

Example 13 Important

Example 16 Important

Ex 7.3,3 Important

Ex 7.3,6 Important

Ex 7.3,10 Important

Example 19 Important

Ex 7.4, 6 Important

Ex 7.4, 8 Important

Example 23 Important You are here

Misc 3 Important

Misc 4 Important

misc 7 Important

Misc 10 Important

Misc 11 Important

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.