      1. Chapter 7 Class 11 Permutations and Combinations
2. Serial order wise
3. Ex 7.3

Transcript

Ex 7.3, 4 (Method 1) Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even? Let the 4 digit number be Total number of digits (1, 2, 3, 4, 5) = 5 So, n = 5 We need to make 4-digit number, so we take 4 digits at a time, So, r = 4 Number of 4-digit numbers = nPr = 5P4 = 5!/(5 − 4)! = 5!/1! = 5! = 5 × 4 × 3 × 2 × 1 = 120 Thus, number of 4-digit numbers = 120 Now, we have to even 4-digit numbers In an even number, last digit should be even So, here units place should be even In our digits (1, 2, 3, 4, 5), only 2,4 are even Hence, either 2 or 4 can be at units place So, we fix 2 at units place and then find the total ways Digit at unit place = 2 Remaining 3 places are to be filled with digit (1, 3, 4, 5) Hence, n = Number of digits left = 4 & r = Number of digits to be taken = 3 Number of ways remaining 3 places are filled = 4P3 = 4!/(4 − 3)! = 4!/1! = 4 × 3 × 2 = 24 Similarly, if we fix digit at unit place as 4, Number of ways = 24 Total number to 4 digit even number = 24 + 24 = 48 Ex7.3, 4 (Method 2) Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even? Let the 4 digit number be Number of 4 digit numbers = 5 × 4 × 3 × 2 = 120 Now, we find 4 digit even numbers Of our digits (1, 2, 3, 4, 5) , only 2 numbers are possible at units place (2 , 4) as we need even number. Number of 3 digit even numbers = 2 × 4 × 3 × 2 = 48

Ex 7.3 